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Physics · Secondary School · Kinematics

Kinematics

A bus pulling up at the stop, a car braking at the lights, a ball kicked at the goal and a boat crossing a river are motions you probably see every day, and Kinematics describes motions like these without asking what caused them. We want to know where an object is, how fast it moves and how that speed changes, and we will play with each of these motions until the formulas make sense.

  1. 1Reference frame and displacement
  2. 2Average velocity
  3. 3Uniform motion
  4. 4When two objects meet
  5. 5Acceleration
  6. 6Free fall
  7. 7Projectiles
  8. 8Adding velocities
  9. ✓Challenges
STEP 1

At rest or moving? It depends on who is watching

To say where something is, we first need to choose a reference frame, the object from which we measure. On a motorway, the position s is the reading on the kilometre markers, as if there were a long ruler laid along the path.

A seated passenger is at rest relative to the bus and, at the same time, moving relative to the person at the bus stop. The path (trajectory) itself depends on the frame, and we can see this with a ball tossed straight up inside the bus, which goes up and down in a straight line for the passenger while, for someone on the pavement, it traces an arc.

Displacement \(\Delta s\) only compares the end with the start, while the distance travelled \(d\) adds up the whole path, including every back-and-forth.

\(\Delta s = s - s_0\)\(d = |\Delta s_1| + |\Delta s_2| + \dots\)\(d \geq |\Delta s|\)\(\Delta s\) can be negative, when the object moves in the direction where the markers decrease, and \(d\) is never negative. Anyone who returns to the starting point has \(\Delta s = 0\).

displacement Δs   legs that add up to the distance d

Let's discuss

  • Choose ‘Out and back’ and pause when the bus turns around at km 8. What are Δs and d at that moment, and what do they become at the end of the trip?
  • When we switch to ‘View from inside the bus’, who starts moving, the bus or the bus stop?
  • With the bus moving, press ‘Toss ball up’ in both frames and compare the path we see for the ball in each one.
  • In ‘Back to start’, the bus travelled 14 km; why, then, does Δs end up at zero?
  • Toss the ball just before the bus stops to turn around. Where does it land, and what could explain that result?
Time
Position s
Displacement Δs
Distance d
Velocity (road frame)
Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A passenger is sitting on a bus travelling at 60 km/h along a straight avenue. Is she at rest or moving?

    Show solution
    The answer depends on the reference frame we choose.
    Relative to the bus (seat, driver), her position does not change, and she is at rest.
    Relative to the street or to a person at the stop, her position changes, and she is moving at \(60\,\text{km/h}\).
    Rest and motion only make sense once we say ‘relative to what’.
  2. basic

    A car passes the km 20 marker on a motorway and, later, the km 70 marker, without turning back. What was its displacement, and what distance did it travel?

    Show solution
    \(\Delta s = s - s_0 = 70 - 20 = 50\,\text{km}\)
    Since the car never reversed, the distance equals the magnitude of the displacement.
    \(\Delta s = 50\,\text{km}\) and \(d = 50\,\text{km}\)
  3. basic

    A bus leaves km 30, goes to km 80 and comes back to km 50. Find the displacement and the distance travelled.

    Show solution
    For the displacement, only the start and the end matter.
    \(\Delta s = 50 - 30 = 20\,\text{km}\)
    For the distance, we add up every leg.
    \(d = (80 - 30) + (80 - 50) = 50 + 30 = 80\,\text{km}\)
    \(\Delta s = 20\,\text{km}\); \(d = 80\,\text{km}\)
  4. basic

    Inside a train moving in a straight line at constant speed, a girl drops a ball. What path does she see? And a person standing on the platform?

    Show solution
    For the girl, she and the ball have the same horizontal velocity, so the ball falls straight down to the floor of the carriage.
    For the person on the platform, the ball falls and, at the same time, moves forward with the train, so the path is an arc of a parabola.
    The path depends on the frame: straight for the girl, curved for someone outside.
  5. intermediate

    An athlete runs 3 full laps of a 400 m track and stops exactly at the starting line. What distance did she travel, and what is her displacement?

    Show solution
    \(d = 3 \cdot 400 = 1200\,\text{m}\)
    She finishes where she started, that is, final position = initial position.
    \(\Delta s = 0\)
    \(d = 1200\,\text{m}\) and \(\Delta s = 0\)
  6. intermediate

    A car passes km 120 at 10:00 and km 90 at 10:30. What was its displacement? What does the sign tell you?

    Show solution
    \(\Delta s = 90 - 120 = -30\,\text{km}\)
    The negative sign means the car moved opposite to the direction in which the marker numbers increase (backward, or retrograde, motion).
    The distance travelled is \(30\,\text{km}\), since distance is never negative.
    \(\Delta s = -30\,\text{km}\): motion towards decreasing markers.
  7. intermediate

    On a road, the origin \((s = 0)\) is the main square of town A, and the positive direction points towards town B. A motorbike starts 15 km before the square and goes to 25 km past it, without turning back. What are \(s_0\), \(s\) and \(\Delta s\)?

    Show solution
    \(s_0 = -15\,\text{km}\) (before the origin) and \(s = +25\,\text{km}\)
    \(\Delta s = 25 - (-15) = 40\,\text{km}\)
    \(s_0 = -15\,\text{km}\), \(s = 25\,\text{km}\), \(\Delta s = 40\,\text{km}\)
  8. intermediate

    Can the distance travelled ever be smaller than the magnitude of the displacement? And in what situation are the two equal?

    Show solution
    Displacement joins the starting point to the finishing point; distance adds up the whole path, including any back-and-forth.
    That is why we always have \(d \geq |\Delta s|\).
    They are equal when the object moves in a straight line without reversing.
    No. \(d = |\Delta s|\) only when there is no turning back.
  9. challenge

    A delivery driver leaves km 10, drives to km 46, turns back to km 28 and then goes on to km 40, where the trip ends. Find the total displacement and the distance travelled.

    Show solution
    \(\Delta s = 40 - 10 = 30\,\text{km}\)
    Legs: \(46 - 10 = 36\), \(46 - 28 = 18\), \(40 - 28 = 12\)
    \(d = 36 + 18 + 12 = 66\,\text{km}\)
    \(\Delta s = 30\,\text{km}\); \(d = 66\,\text{km}\)
  10. challenge

    A passenger walks 20 m from the back to the front of a carriage in 10 s. In that time, the train moves 150 m relative to the tracks. What is the passenger’s displacement relative to the carriage? And relative to the tracks? What if he had walked from the front to the back?

    Show solution
    Relative to the carriage, \(\Delta s = 20\,\text{m}\).
    Relative to the tracks, the displacements add, since they point the same way: \(150 + 20 = 170\,\text{m}\).
    Walking from front to back, he moves \(150 - 20 = 130\,\text{m}\) forward relative to the tracks.
    20 m (carriage); 170 m (tracks); 130 m if he walked backward.
STEP 2

Average velocity: how far you went divided by how long it took

Average velocity compares the displacement with the time taken, and that time includes everything, from stops to traffic lights and traffic jams. The speedometer measures another quantity, the instantaneous velocity, which is the velocity at each moment.

It seems natural to average the speeds on each leg, but average velocity is not that average, because driving slowly means spending more time on that leg and, for that reason, the slow leg ‘weighs’ more in the result.

On the \(s \times t\) graph, the average velocity appears as the slope of the line joining the starting point to the current point.

\(v_m = \dfrac{\Delta s}{\Delta t}\)\(1\,\text{m/s} = 3.6\,\text{km/h}\)\(\text{km/h} \xrightarrow{\;\div\, 3.6\;} \text{m/s}\)In SI units, velocity is given in m/s. It is worth memorising a few pairs, such as 36 km/h = 10 m/s · 72 km/h = 20 m/s · 108 km/h = 30 m/s.

Let's discuss

  • Choose ‘Half at 60, half at 40’ and place your bet before running it. Our first intuition tends to be 50 km/h; check at the end whether the average really comes out that way.
  • Increase the stop to 60 min and watch what the graph looks like during the stop. What happens, then, to the average velocity?
  • Over the trip, at which moments does the speedometer read higher than \(v_m\), and at which does it read lower?
  • Convert the speedometer reading to m/s in your head and check the result against the reading in m/s.
  • When does the average of the legs match the average velocity? It may be worth testing a hunch in ‘Always at 80’.
Stopwatch
Covered so far
Speedometer
Speedometer in m/s
vₘ so far
Average of the legs
Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A car is travelling at 72 km/h. What is that in m/s?

    Show solution
    Since \(1\,\text{m/s} = 3.6\,\text{km/h}\), we just divide by 3.6.
    \(v = \dfrac{72}{3.6} = 20\,\text{m/s}\)
    \(v = 20\,\text{m/s}\)
  2. basic

    During a trip, a car’s speedometer showed 0, 40, 100 and 60 km/h at different moments. Is the average velocity for the trip the average of these numbers? What does the speedometer measure?

    Show solution
    The speedometer measures instantaneous speed, that is, how fast the car is going at that moment.
    Average velocity is \(v_m = \dfrac{\Delta s}{\Delta t}\), which depends on how far the car went and how much time passed, stops included.
    If we average a few speedometer readings, we ignore how long the car spent at each speed.
    No. The speedometer gives the instantaneous speed; the average comes from \(\Delta s / \Delta t\).
  3. basic

    A family drives 180 km in 2 h 30 min. What is the average velocity for the trip?

    Show solution
    \(\Delta t = 2.5\,\text{h}\)
    \(v_m = \dfrac{\Delta s}{\Delta t} = \dfrac{180}{2.5} = 72\,\text{km/h}\)
    \(v_m = 72\,\text{km/h}\)
  4. basic

    Usain Bolt ran 100 m in 9.58 s. What was his average velocity in m/s and in km/h?

    Show solution
    \(v_m = \dfrac{100}{9.58} \approx 10.44\,\text{m/s}\)
    \(10.44 \cdot 3.6 \approx 37.6\,\text{km/h}\)
    \(v_m \approx 10.4\,\text{m/s} \approx 37.6\,\text{km/h}\)
  5. intermediate

    On a trip, a car covers 120 km in 1.5 h, stops for 30 min at a petrol station and then covers another 80 km in 1 h. What is the average velocity for the whole trip?

    Show solution
    \(\Delta s = 120 + 80 = 200\,\text{km}\)
    \(\Delta t = 1.5 + 0.5 + 1 = 3\,\text{h}\) (the stop counts!)
    \(v_m = \dfrac{200}{3} \approx 66.7\,\text{km/h}\)
    \(v_m \approx 66.7\,\text{km/h}\)
  6. intermediate

    A car covers 120 km, the first half of the way at 60 km/h and the second half at 40 km/h. Is the average velocity 50 km/h? Do the calculation before answering.

    Show solution
    1st half: \(\Delta t_1 = \dfrac{60}{60} = 1\,\text{h}\)
    2nd half: \(\Delta t_2 = \dfrac{60}{40} = 1.5\,\text{h}\)
    \(v_m = \dfrac{120}{1 + 1.5} = 48\,\text{km/h}\)
    Since the car spends more time on the slow leg, the average ends up below 50.
    No: \(v_m = 48\,\text{km/h}\).
  7. intermediate

    On a 10 km stretch monitored by average-speed cameras, the limit is 80 km/h. A car takes 6 min to cover the stretch. Will it get a ticket?

    Show solution
    \(\Delta t = 6\,\text{min} = 0.1\,\text{h}\)
    \(v_m = \dfrac{10}{0.1} = 100\,\text{km/h}\)
    To respect the limit, it should take at least \(\dfrac{10}{80} = 0.125\,\text{h} = 7.5\,\text{min}\).
    Yes: \(v_m = 100\,\text{km/h} > 80\,\text{km/h}\).
  8. intermediate

    Sunlight travels about \(1.5 \cdot 10^{11}\,\text{m}\) to reach Earth, at \(3 \cdot 10^{8}\,\text{m/s}\). How long does it take?

    Show solution
    \(\Delta t = \dfrac{\Delta s}{v}\) \(= \dfrac{1.5 \cdot 10^{11}}{3 \cdot 10^{8}}\) \(= 500\,\text{s}\)
    \(500\,\text{s} = 8\,\text{min}\;20\,\text{s}\)
    We see the Sun as it was a little over 8 minutes ago.
    \(\Delta t = 500\,\text{s} \approx 8.3\,\text{min}\)
  9. challenge

    A car goes from A to B at 60 km/h and returns from B to A, along the same road, at 40 km/h. The distance AB is 120 km. Find the average speed for the whole trip (distance/time) and the average velocity (displacement/time).

    Show solution
    Out: \(\dfrac{120}{60} = 2\,\text{h}\); back: \(\dfrac{120}{40} = 3\,\text{h}\)
    Average speed: \(\dfrac{240}{5} = 48\,\text{km/h}\)
    The car returned to A, so the total displacement is zero, and so is the average velocity.
    48 km/h (speed) and 0 (velocity).
  10. challenge

    A car drives the first half of the trip time at 80 km/h and the second half of the time at 40 km/h. What is the average velocity? Compare with the case where it drives half of the distance at each speed.

    Show solution
    With half the time at each speed, we call each half \(T\).
    \(v_m = \dfrac{80T + 40T}{2T} = 60\,\text{km/h}\) (here the simple average does work).
    With half the distance at each speed, we call each half \(D\).
    \(v_m = \dfrac{2D}{\dfrac{D}{80} + \dfrac{D}{40}}\) \(= \dfrac{2D}{\dfrac{3D}{80}}\) \(= \dfrac{160}{3} \approx 53.3\,\text{km/h}\)
    60 km/h by time; ≈ 53.3 km/h by distance.
STEP 3

Uniform motion: the same velocity all the time

In uniform motion along a straight line, the velocity is constant, and the object covers equal distances in equal times. The expression that gives its position at each instant is called the position–time equation.

On the \(s \times t\) graph, uniform motion appears as a straight line whose slope is the velocity, while on the \(v \times t\) graph it becomes a horizontal line, and the area between it and the time axis is the displacement.

When \(v > 0\), positions increase and the motion is forward (progressive); when \(v < 0\), they decrease and the motion is backward (retrograde).

\(s = s_0 + v\,t\)\(v = \dfrac{\Delta s}{\Delta t}\) (slope of \(s \times t\))\(\Delta s = \text{area under } v \times t\)The area below the time axis counts as negative, because it corresponds to a displacement in the direction where positions decrease.

Let's discuss

  • Set \(v = 0\) and describe what the two graphs look like in that case.
  • Choose ‘Backward’. Does the \(s \times t\) line go up or down, and on which side of the axis is the \(v \times t\) area?
  • Change only \(s_0\) and see what happens to the line; does the slope change too?
  • Pause near \(t = 4\,\text{s}\) and check whether the area under \(v \times t\) is equal to \(s - s_0\).
  • If we double \(v\), what happens to the slope and to the area, compared with before?
Time
Position s
Displacement s − s₀
Area under v × t
Type
Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    An object’s position–time equation is \(s = 20 + 5t\) (SI units). What are the initial position, the velocity and the position at \(t = 4\,\text{s}\)?

    Show solution
    Comparing with \(s = s_0 + v\,t\): \(s_0 = 20\,\text{m}\) and \(v = 5\,\text{m/s}\).
    \(s(4) = 20 + 5 \cdot 4 = 40\,\text{m}\)
    \(s_0 = 20\,\text{m}\); \(v = 5\,\text{m/s}\); \(s(4) = 40\,\text{m}\)
  2. basic

    On the \(s \times t\) graph of a uniform motion, the line passes through (0 s, 10 m) and (5 s, 40 m). What are the velocity and the position–time equation?

    Show solution
    The velocity is the slope of the line, which we compute from two points:
    \(v = \dfrac{\Delta s}{\Delta t} = \dfrac{40 - 10}{5 - 0} = 6\,\text{m/s}\)
    \(s = 10 + 6t\) (SI)
  3. basic

    An object follows \(s = 100 - 8t\) (SI). Is the motion forward or backward? At what instant does it pass the origin?

    Show solution
    Since \(v = -8\,\text{m/s}\) is negative, positions decrease and the motion is backward (retrograde).
    Origin: \(0 = 100 - 8t\) \(\Rightarrow t = 12.5\,\text{s}\)
    Backward; it passes the origin at \(t = 12.5\,\text{s}\).
  4. basic

    On the \(v \times t\) graph of a uniform motion, what does the area under the line represent? Calculate it for \(v = 15\,\text{m/s}\) over 8 s.

    Show solution
    The area is a rectangle, and base × height \(= \Delta t \cdot v = \Delta s\), which is the displacement.
    \(\Delta s = 15 \cdot 8 = 120\,\text{m}\)
    The area is the displacement: 120 m.
  5. intermediate

    A 200 m long train, moving at a constant 20 m/s, crosses a 400 m bridge. How long does it take from the moment the front enters the bridge until the back leaves it?

    Show solution
    The front has to cover the whole bridge plus the length of the train.
    \(\Delta s = 400 + 200 = 600\,\text{m}\)
    \(\Delta t = \dfrac{600}{20} = 30\,\text{s}\)
    \(\Delta t = 30\,\text{s}\)
  6. intermediate

    A car in uniform motion passes km 40 at 8:00 and km 130 at 9:30. Write its position–time equation (t in hours, counted from 8:00) and say where it will be at 10:00.

    Show solution
    \(v = \dfrac{130 - 40}{1.5} = 60\,\text{km/h}\)
    \(s = 40 + 60t\)
    At 10:00, \(t = 2\,\text{h}\): \(s = 40 + 60 \cdot 2 = 160\,\text{km}\)
    \(s = 40 + 60t\); at 10:00, at km 160.
  7. intermediate

    On an \(s \times t\) graph, two cars in uniform motion appear as parallel lines, one above the other. What does that tell you about their velocities and positions? Do they ever meet?

    Show solution
    Parallel lines have the same slope, so the cars have the same velocity.
    Since one is above the other, their positions differ at every instant.
    The distance between them never changes, so the lines never cross.
    Same velocity, different positions; they never meet.
  8. intermediate

    A car moves at 20 m/s for 10 s and then goes back at 10 m/s for 6 s (velocity \(-10\,\text{m/s}\)). Using the areas on the \(v \times t\) graph, find the total displacement and the distance travelled.

    Show solution
    Area above the time axis: \(20 \cdot 10 = 200\,\text{m}\)
    Area below the axis: \(-10 \cdot 6 = -60\,\text{m}\)
    \(\Delta s = 200 - 60 = 140\,\text{m}\)
    \(d = 200 + 60 = 260\,\text{m}\)
    \(\Delta s = 140\,\text{m}\); \(d = 260\,\text{m}\)
  9. challenge

    A person shouts in front of a cliff wall and hears the echo 1.5 s later. The speed of sound in air is 340 m/s. How far away is the wall?

    Show solution
    The sound travels to the wall and back, covering twice the distance.
    \(2D = 340 \cdot 1.5 = 510\,\text{m}\)
    \(D = 255\,\text{m}\)
  10. challenge

    A bus in uniform motion is at position 26 m at \(t = 2\,\text{s}\) and at position 6 m at \(t = 7\,\text{s}\). Find its position–time equation and the instant it passes the origin.

    Show solution
    \(v = \dfrac{6 - 26}{7 - 2} = -4\,\text{m/s}\)
    \(s_0 = s - v\,t = 26 - (-4) \cdot 2 = 34\,\text{m}\)
    \(s = 34 - 4t\)
    Origin: \(0 = 34 - 4t\) \(\Rightarrow t = 8.5\,\text{s}\)
    \(s = 34 - 4t\) (SI); it passes the origin at \(8.5\,\text{s}\).
STEP 4

When and where do two objects meet?

Two objects meet when they are at the same position at the same instant, and to find that instant we write the position–time equation for each one, using the same origin and the same positive direction, and set them equal.

On the \(s \times t\) graph, the meeting shows up as the point where the lines cross, whose horizontal coordinate gives the instant and whose vertical coordinate gives the position.

There is a shortcut that can save us a lot of algebra. In opposite directions the speeds add, because the objects close in faster, and in the same direction they subtract, since the faster one has to ‘close the gap’, which is what matters when overtaking.

\(s_A = s_B\)\(s_{0A} + v_A\,t = s_{0B} + v_B\,t\)\(t_E = \dfrac{s_{0B} - s_{0A}}{v_A - v_B}\)If \(v_A = v_B\), the lines are parallel and there is no meeting, unless the objects start together. A negative \(t_E\) means the meeting already happened in the past.

car A   car B   meeting point

Let's discuss

  • In ‘Opposite directions’, work out \(t_E\) and the meeting position before running and, afterwards, check the result on the graph.
  • In ‘Overtaking’, reduce the speed difference between A and B and see whether the meeting now happens sooner or later.
  • In ‘Side by side’, the velocities are equal; why, in that case, do the lines never cross?
  • In ‘Moving apart’, the calculation gives a negative time. What could that negative time mean on the road?
Time
Position of A
Position of B
Distance between them
Meeting time
Meeting position
Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    Two objects follow \(s_A = 10 + 5t\) and \(s_B = 70 - 10t\) (SI). When and where do they meet?

    Show solution
    At the meeting, \(s_A = s_B\)
    \(10 + 5t = 70 - 10t\) \(\Rightarrow 15t = 60\) \(\Rightarrow t = 4\,\text{s}\)
    \(s = 10 + 5 \cdot 4 = 30\,\text{m}\)
    \(t = 4\,\text{s}\), at \(s = 30\,\text{m}\)
  2. basic

    Two cars travel in the same direction at the same velocity, one 50 m behind the other. Does the one behind catch up? What does the \(s \times t\) graph look like?

    Show solution
    With the same velocity, the distance between the two cars always stays 50 m.
    On the \(s \times t\) graph, the lines are parallel and never cross.
    They never meet; that would only happen if the car behind were faster.
  3. basic

    Two cars are 300 m apart and drive towards each other at 20 m/s and 10 m/s. After how long do they meet? How far from the faster car’s starting point?

    Show solution
    Placing the origin at the faster car, \(s_A = 20t\) and \(s_B = 300 - 10t\)
    \(20t = 300 - 10t\) \(\Rightarrow t = 10\,\text{s}\)
    \(s = 20 \cdot 10 = 200\,\text{m}\)
    With the shortcut, since in opposite directions the speeds add, \(t = \dfrac{300}{20 + 10} = 10\,\text{s}\).
    \(t = 10\,\text{s}\), 200 m from the faster car’s start.
  4. basic

    Car A, at position 0 and moving at 25 m/s, chases car B, which is at position 100 m and moving at 15 m/s in the same direction. When and where does A catch B?

    Show solution
    \(25t = 100 + 15t\) \(\Rightarrow 10t = 100\) \(\Rightarrow t = 10\,\text{s}\)
    \(s = 25 \cdot 10 = 250\,\text{m}\)
    With the shortcut, in the same direction A closes the gap at \(25 - 15 = 10\,\text{m/s}\).
    \(t = 10\,\text{s}\), at \(s = 250\,\text{m}\)
  5. intermediate

    Towns A and B are 240 km apart. At the same time, a car leaves A for B at 80 km/h and a motorbike leaves B for A at 40 km/h. When, and how far from A, do they pass each other?

    Show solution
    \(s_{\text{car}} = 80t\) and \(s_{\text{bike}} = 240 - 40t\)
    \(80t = 240 - 40t\) \(\Rightarrow 120t = 240\) \(\Rightarrow t = 2\,\text{h}\)
    \(s = 80 \cdot 2 = 160\,\text{km}\)
    After 2 h, 160 km from A.
  6. intermediate

    A 4 m car at 26 m/s is going to overtake a 20 m lorry at 20 m/s. The overtaking starts when the front of the car reaches the back of the lorry and ends when the back of the car passes the front of the lorry. How long does it last? How far does the car travel relative to the road?

    Show solution
    Relative to the lorry, the car has to move ahead \(20 + 4 = 24\,\text{m}\).
    Relative velocity: \(26 - 20 = 6\,\text{m/s}\)
    \(\Delta t = \dfrac{24}{6} = 4\,\text{s}\)
    Relative to the road: \(26 \cdot 4 = 104\,\text{m}\)
    4 s; the car travels 104 m.
  7. intermediate

    At 8:00, a lorry leaves km 0 at 60 km/h. At 9:00, a car leaves the same point, in the same direction, at 90 km/h. At what time, and at which km, does the car catch the lorry?

    Show solution
    With t in hours from 8:00: \(s_{\text{lorry}} = 60t\) and \(s_{\text{car}} = 90(t - 1)\)
    \(60t = 90t - 90\) \(\Rightarrow 30t = 90\) \(\Rightarrow t = 3\,\text{h}\)
    \(s = 60 \cdot 3 = 180\,\text{km}\)
    At 11:00, at km 180.
  8. intermediate

    Two trains run on parallel tracks in opposite directions: one is 150 m long and moves at 20 m/s; the other is 100 m long and moves at 30 m/s. How long does the complete crossing last (from when the fronts meet until the backs separate)?

    Show solution
    Relative velocity (opposite directions): \(20 + 30 = 50\,\text{m/s}\)
    Length to cover: \(150 + 100 = 250\,\text{m}\)
    \(\Delta t = \dfrac{250}{50} = 5\,\text{s}\)
    \(\Delta t = 5\,\text{s}\)
  9. challenge

    Two trains are 100 km apart on the same line, heading towards each other at 50 km/h each. A bird flies at 75 km/h from one train to the other and back, non-stop, until the trains meet. How many kilometres does the bird fly?

    Show solution
    We do not need to add up all the back-and-forth trips; knowing the time is enough.
    Trains meet at: \(t = \dfrac{100}{50 + 50} = 1\,\text{h}\)
    Since the bird flies the whole time at 75 km/h, \(d = 75 \cdot 1 = 75\,\text{km}\)
    \(d = 75\,\text{km}\)
  10. challenge

    On the same straight road, cyclist A is at \(s = 0\) moving at 18 km/h and cyclist B is at \(s = 1200\,\text{m}\) moving at 2 m/s, in the same direction. When and where does A catch B? How far apart are they 10 min after meeting?

    Show solution
    Converting to the same units, \(18\,\text{km/h} = \dfrac{18}{3.6} = 5\,\text{m/s}\)
    \(5t = 1200 + 2t\) \(\Rightarrow 3t = 1200\) \(\Rightarrow t = 400\,\text{s}\) (6 min 40 s)
    \(s = 5 \cdot 400 = 2000\,\text{m}\)
    After meeting, A pulls away at \(5 - 2 = 3\,\text{m/s}\): in \(600\,\text{s}\), \(3 \cdot 600 = 1800\,\text{m}\)
    At 400 s, at the 2000 m point; 10 min later they are 1800 m apart.
STEP 5

Acceleration: when the velocity changes

Acceleration tells you how much the velocity changes every second; with \(a = 3\,\text{m/s}^2\), for example, the velocity gains 3 m/s each second. When the acceleration is constant, we call the motion uniformly accelerated motion.

In this motion, the velocity changes linearly (\(v \times t\) is a straight line) and the position follows a parabola (\(s \times t\)). If the position is marked every second, the marks get farther apart when the car speeds up and closer together when it brakes.

Torricelli’s equation (the v² equation) links velocity and displacement without using time, and it is the go-to formula for braking distances.

\(a = \dfrac{\Delta v}{\Delta t}\)\(v = v_0 + a\,t\)\(s = s_0 + v_0\,t + \dfrac{a\,t^2}{2}\)\(v^2 = v_0^2 + 2\,a\,\Delta s\)\(a\) in m/s². The motion speeds up when \(v\) and \(a\) have the same sign and slows down when they have opposite signs. The slope of \(v \times t\) gives \(a\), and the area under \(v \times t\) gives \(\Delta s\).

position   velocity   acceleration   marks every 1 s

Let's discuss

  • In ‘From rest’, read the distances between the 1 s, 2 s, 3 s… marks and say what sequence of numbers seems to emerge.
  • In ‘Braking’, at what instant does the car stop? We can check the instant with \(v = v_0 + a\,t\), and the distance with Torricelli.
  • Turn off ‘real brakes’ and see what the formula would do after \(v = 0\), something that may surprise you.
  • Choose ‘Reversing, speeding up’, in which \(a\) is negative, and decide whether the car is braking.
  • Compare the readings \(v^2\) and \(v_0^2 + 2a\Delta s\) at several instants. What does this comparison suggest regarding Torricelli’s equation?
Time
Position s
Velocity v
Acceleration a
v² now
v₀² + 2aΔs
Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A car goes from 0 to 108 km/h in 10 s. What is its average acceleration in \(\text{m/s}^2\)?

    Show solution
    \(108\,\text{km/h} = \dfrac{108}{3.6} = 30\,\text{m/s}\)
    \(a = \dfrac{\Delta v}{\Delta t} = \dfrac{30 - 0}{10} = 3\,\text{m/s}^2\)
    Every second, the velocity increases by 3 m/s.
    \(a = 3\,\text{m/s}^2\)
  2. basic

    An object’s velocity is \(v = 5 + 2t\) (SI). What are \(v_0\), \(a\) and the velocity at \(t = 6\,\text{s}\)?

    Show solution
    Comparing with \(v = v_0 + a\,t\): \(v_0 = 5\,\text{m/s}\) and \(a = 2\,\text{m/s}^2\).
    \(v(6) = 5 + 2 \cdot 6 = 17\,\text{m/s}\)
    \(v_0 = 5\,\text{m/s}\); \(a = 2\,\text{m/s}^2\); \(v(6) = 17\,\text{m/s}\)
  3. basic

    An object’s position is \(s = 4 + 3t + 2t^2\) (SI). Identify \(s_0\), \(v_0\) and \(a\), and calculate \(s\) at \(t = 2\,\text{s}\).

    Show solution
    Comparing with \(s = s_0 + v_0\,t + \dfrac{a}{2}\,t^2\): \(s_0 = 4\,\text{m}\), \(v_0 = 3\,\text{m/s}\) and \(\dfrac{a}{2} = 2\) \(\Rightarrow a = 4\,\text{m/s}^2\).
    \(s(2) = 4 + 3 \cdot 2 + 2 \cdot 2^2 = 4 + 6 + 8 = 18\,\text{m}\)
    \(s_0 = 4\,\text{m}\); \(v_0 = 3\,\text{m/s}\); \(a = 4\,\text{m/s}^2\); \(s(2) = 18\,\text{m}\)
  4. basic

    Does negative acceleration always mean the car is braking? Explain with an example.

    Show solution
    No. What decides is the comparison between the signs of \(v\) and \(a\).
    With the same signs, the speed (magnitude of velocity) increases (speeding up); with opposite signs, it decreases (slowing down).
    For example, with \(v = -10\,\text{m/s}\) and \(a = -2\,\text{m/s}^2\), the car moves in the negative direction faster and faster.
    Negative acceleration with negative velocity means speeding up.
  5. intermediate

    A car at 20 m/s brakes with a constant acceleration of magnitude \(5\,\text{m/s}^2\). What distance does it cover before stopping, and how long does it take?

    Show solution
    Torricelli: \(0 = 20^2 + 2 \cdot (-5) \cdot \Delta s\) \(\Rightarrow \Delta s = \dfrac{400}{10} = 40\,\text{m}\)
    Time: \(0 = 20 - 5t\) \(\Rightarrow t = 4\,\text{s}\)
    40 m in 4 s.
  6. intermediate

    On a \(v \times t\) graph, the velocity goes from 4 m/s at \(t = 0\) to 16 m/s at \(t = 6\,\text{s}\) along a straight line. Find the acceleration and the displacement.

    Show solution
    The acceleration is the slope: \(a = \dfrac{16 - 4}{6} = 2\,\text{m/s}^2\)
    The displacement is the area of the trapezoid: \(\Delta s = \dfrac{(4 + 16) \cdot 6}{2} = 60\,\text{m}\)
    \(a = 2\,\text{m/s}^2\); \(\Delta s = 60\,\text{m}\)
  7. intermediate

    An object starts from rest with \(a = 2\,\text{m/s}^2\). How far does it travel during the 1st, 2nd and 3rd second? What pattern appears?

    Show solution
    With \(s_0 = 0\) and \(v_0 = 0\): \(s = \dfrac{2}{2}\,t^2 = t^2\).
    \(s(1) = 1\,\text{m}\), \(s(2) = 4\,\text{m}\), \(s(3) = 9\,\text{m}\)
    1st second: 1 m; 2nd: \(4 - 1 = 3\,\text{m}\); 3rd: \(9 - 4 = 5\,\text{m}\)
    1, 3, 5 m: the odd numbers (Galileo’s rule). The marks get farther and farther apart.
  8. intermediate

    A plane needs to reach 80 m/s to take off and accelerates at \(4\,\text{m/s}^2\) from rest. What is the minimum runway length? How long does it take?

    Show solution
    \(v^2 = v_0^2 + 2a\,\Delta s\) \(\Rightarrow 80^2 = 0 + 2 \cdot 4 \cdot \Delta s\) \(\Rightarrow 6400 = 8\,\Delta s\) \(\Rightarrow \Delta s = 800\,\text{m}\)
    \(t = \dfrac{80}{4} = 20\,\text{s}\)
    800 m of runway; 20 s.
  9. challenge

    A driver at 72 km/h sees an obstacle 50 m ahead. It takes him 0.7 s to react (during that time the car keeps moving uniformly) and then he brakes at \(5\,\text{m/s}^2\). (a) Does he stop before the obstacle? (b) What is the highest speed that would let him stop in time?

    Show solution
    (a) \(72\,\text{km/h} = 20\,\text{m/s}\). Reaction: \(20 \cdot 0.7 = 14\,\text{m}\). Braking: \(\dfrac{20^2}{2 \cdot 5} = 40\,\text{m}\). Total: \(54\,\text{m} > 50\,\text{m}\). He does not stop: he crashes.
    (b) \(0.7\,v + \dfrac{v^2}{10} = 50\) \(\Rightarrow v^2 + 7v - 500 = 0\)
    \(v = \dfrac{-7 + \sqrt{49 + 2000}}{2} \approx 19.1\,\text{m/s} \approx 68.9\,\text{km/h}\)
    (a) No (he would need 54 m). (b) About 19.1 m/s ≈ 69 km/h.
  10. challenge

    Car A starts from rest with \(a = 2\,\text{m/s}^2\) at the exact instant car B passes it at a constant 10 m/s. When and where does A catch B? What is A’s velocity at that moment?

    Show solution
    \(s_A = \dfrac{2}{2}\,t^2 = t^2\) and \(s_B = 10t\)
    \(t^2 = 10t\) \(\Rightarrow t = 10\,\text{s}\) (\(t = 0\) is the start)
    \(s = 10 \cdot 10 = 100\,\text{m}\)
    \(v_A = 2 \cdot 10 = 20\,\text{m/s}\), twice B’s velocity.
    At 10 s, 100 m away; A is moving at 20 m/s.
STEP 6

Free fall: without air, everything falls together

Near the Earth’s surface, and assuming there is no air resistance, every object falls with the same acceleration, \(g \approx 10\,\text{m/s}^2\), whatever its mass. On the Moon, which has no air, an astronaut dropped a hammer and a feather, and the two hit the ground together.

Free fall is uniformly accelerated motion with \(a = g\) pointing down. When we throw an object upward, it loses 10 m/s every second while rising, stops for an instant at the top (\(v = 0\), but the acceleration is still \(g\)!) and comes back down, taking, if we neglect air resistance, the same time to fall as it took to rise.

\(v = v_0 - g\,t\)\(y = y_0 + v_0\,t - \dfrac{g\,t^2}{2}\)\(t_q = \sqrt{\dfrac{2h}{g}}\)\(H = \dfrac{v_0^2}{2g}\)We take the \(y\) axis pointing up and \(g = 10\,\text{m/s}^2\). An object dropped from rest at height \(h\) reaches the ground with \(v = \sqrt{2gh}\).

velocity   acceleration g

Let's discuss

  • Drop it from 45 m and try to estimate how long the ball takes to reach the ground, then check with \(t = \sqrt{2h/g}\).
  • Turn on slow motion and watch the gaps between the snapshots, taken every 0.1 s. Do they seem to grow in the ratio 1, 3, 5…?
  • Throw it up at 20 m/s, pause at the top and see what the velocity and the acceleration are at that instant.
  • Switch to ‘With air’ and see which of the two reaches the ground first, the feather or the hammer. And what changes when we remove the air?
Time
Ball height
Velocity
Acceleration
Time to the ground
Maximum height
Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A stone is dropped from the top of a 20 m wall. How long does it take to reach the ground, and at what speed does it arrive? Use \(g = 10\,\text{m/s}^2\) and ignore air resistance.

    Show solution
    \(t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 \cdot 20}{10}} = \sqrt{4} = 2\,\text{s}\)
    \(v = g\,t = 10 \cdot 2 = 20\,\text{m/s}\)
    2 s; 20 m/s.
  2. basic

    In 1971, an astronaut on the Moon dropped a hammer and a feather at the same time. What happened, and why would it be different on Earth?

    Show solution
    Since there is no air on the Moon, both have the same acceleration (lunar gravity) and hit the ground together.
    On Earth, air resistance slows the feather much more, because it is light and wide.
    Without air, falling does not depend on mass; the everyday difference comes from the air.
  3. basic

    A ball is dropped and falls freely for 3 s. What is its velocity, and how far has it fallen? Use \(g = 10\,\text{m/s}^2\).

    Show solution
    \(v = g\,t = 10 \cdot 3 = 30\,\text{m/s}\)
    \(h = \dfrac{g\,t^2}{2} = \dfrac{10 \cdot 9}{2} = 45\,\text{m}\)
    30 m/s; 45 m.
  4. basic

    A ball is thrown upward. At the highest point, its velocity is zero. Is the acceleration also zero at that point?

    Show solution
    No. The acceleration is \(g = 10\,\text{m/s}^2\) downward during the whole flight.
    If the acceleration were zero at the top, the ball would stay hanging there in the air!
    At the top, the velocity is changing from positive (going up) to negative (coming down).
    At the top, \(v = 0\), but \(a = g\) downward.
  5. intermediate

    A ball is thrown upward at 20 m/s. Find the time to rise, the maximum height and the time until it returns to the hand. Use \(g = 10\,\text{m/s}^2\).

    Show solution
    Rising: \(0 = 20 - 10t\) \(\Rightarrow t_s = 2\,\text{s}\)
    Maximum height: \(H = \dfrac{v_0^2}{2g} = \dfrac{400}{20} = 20\,\text{m}\)
    The fall takes the same time as the rise: total \(= 4\,\text{s}\), and the ball comes back at 20 m/s.
    2 s; 20 m; 4 s.
  6. intermediate

    A strobe photo captures a dropped ball every 0.1 s. How far does it fall in the 1st, 2nd and 3rd interval of 0.1 s? Use \(g = 10\,\text{m/s}^2\).

    Show solution
    \(h = \dfrac{g\,t^2}{2} = 5t^2\): \(h(0.1) = 0.05\,\text{m}\), \(h(0.2) = 0.20\,\text{m}\), \(h(0.3) = 0.45\,\text{m}\)
    Intervals: \(5\,\text{cm}\); \(20 - 5 = 15\,\text{cm}\); \(45 - 20 = 25\,\text{cm}\)
    5, 15 and 25 cm: they grow in the ratio 1, 3, 5.
  7. intermediate

    A coin falls from the top of an 80 m building. Neglecting air resistance, how long does it take, and at what speed does it hit the ground (in km/h)? Use \(g = 10\,\text{m/s}^2\).

    Show solution
    \(t = \sqrt{\dfrac{2 \cdot 80}{10}} = \sqrt{16} = 4\,\text{s}\)
    \(v = 10 \cdot 4 = 40\,\text{m/s} = 40 \cdot 3.6 = 144\,\text{km/h}\)
    In reality, the air reduces this value a lot.
    4 s; 144 km/h.
  8. intermediate

    A ball thrown straight up reaches a maximum height of 45 m. At what speed was it thrown, and how long does it stay in the air? Use \(g = 10\,\text{m/s}^2\).

    Show solution
    At the top \(v = 0\): \(0 = v_0^2 - 2gH\) \(\Rightarrow v_0^2 = 2 \cdot 10 \cdot 45 = 900\) \(\Rightarrow v_0 = 30\,\text{m/s}\)
    Rising: \(\dfrac{30}{10} = 3\,\text{s}\); in the air: \(2 \cdot 3 = 6\,\text{s}\)
    30 m/s; 6 s in the air.
  9. challenge

    From the top of a 20 m building, a ball is thrown upward at 15 m/s. How long does it take to reach the ground, and at what speed does it arrive? Use \(g = 10\,\text{m/s}^2\).

    Show solution
    With the axis pointing up and the origin at the top of the building, \(y = 15t - 5t^2\). The ground is at \(y = -20\,\text{m}\).
    \(-20 = 15t - 5t^2\) \(\Rightarrow t^2 - 3t - 4 = 0\) \(\Rightarrow (t - 4)(t + 1) = 0\) \(\Rightarrow t = 4\,\text{s}\)
    \(v = 15 - 10 \cdot 4 = -25\,\text{m/s}\) (25 m/s downward)
    Check with Torricelli: \(v^2 = 15^2 + 2 \cdot 10 \cdot 20 = 625\) \(\Rightarrow |v| = 25\,\text{m/s}\)
    4 s; 25 m/s downward.
  10. challenge

    Ball A is dropped from a height of 45 m. At the same instant, on the same vertical line, ball B is thrown upward from the ground at 30 m/s. When and at what height do they meet? Use \(g = 10\,\text{m/s}^2\).

    Show solution
    \(y_A = 45 - 5t^2\) and \(y_B = 30t - 5t^2\)
    Setting them equal: \(45 = 30t\) \(\Rightarrow t = 1.5\,\text{s}\). The \(5t^2\) term cancels, and we can read this as both ‘falling together’, with the gap between them shrinking at 30 m/s.
    \(y = 45 - 5 \cdot 1.5^2 = 45 - 11.25 = 33.75\,\text{m}\)
    At 1.5 s, 33.75 m above the ground.
STEP 7

Projectiles: two motions at the same time

If we neglect air resistance, a projectile performs two independent motions. Horizontally there is no acceleration, which gives uniform motion with constant \(v_x\); vertically the acceleration \(g\) acts, producing uniformly accelerated motion just like the one we studied in step 6. Together, the two trace a parabola.

The shadows can help us see this independence. The shadow on the ground moves in equal steps in equal times, while the shadow on the wall goes up and down like a ball thrown straight up.

In a horizontal launch (\(\theta = 0^\circ\)), the time to fall is the same as when we simply drop the ball, and it depends only on the height. If the ball launches and lands at the same height, 45° gives the longest range, and complementary angles, such as 30° and 60°, give the same range.

\(v_x = v_0\cos\theta\)\(v_y = v_0\sin\theta - g\,t\)\(x = v_x\,t\)\(H = \dfrac{(v_0\sin\theta)^2}{2g}\)\(A = \dfrac{v_0^2\sin(2\theta)}{g}\)The formulas for \(H\) and the range \(A\) apply to a launch from the ground (launch and landing at the same height), without air, with \(g = 10\,\text{m/s}^2\).

\(v_x\) and shadow on the ground   \(v_y\) and shadow on the wall   \(\vec v\)

Let's discuss

  • Fix \(v_0\) and look for the angle that seems to give the longest range. Is it really 45°?
  • Turn on ‘complementary angle’ and see whether the two shots land in the same place. Which of them goes higher, and which stays longer in the air?
  • During the flight, what happens to \(v_x\) and to \(v_y\)? Pause at the top and see what \(v_y\) is there.
  • In ‘Horizontal off a cliff’, double \(v_0\) and watch whether the fall time changes. What happens to the range?
  • Why are the ground-shadow marks equally spaced while the wall-shadow marks are not?
Time
Horizontal velocity vx
Vertical velocity vy
Range
Maximum height
Flight time
Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A marble rolls across a 0.8 m high table and leaves the edge at 3 m/s. How long does it take to reach the floor, and how far from the table does it land? Use \(g = 10\,\text{m/s}^2\).

    Show solution
    Vertical (free fall): \(t = \sqrt{\dfrac{2 \cdot 0.8}{10}} = \sqrt{0.16} = 0.4\,\text{s}\)
    Horizontal (uniform motion): \(x = 3 \cdot 0.4 = 1.2\,\text{m}\)
    0.4 s; it lands 1.2 m from the table.
  2. basic

    From the same point and at the same instant, one bullet is fired horizontally and another is simply dropped. Which reaches the ground first (no air, flat ground)?

    Show solution
    The vertical motion of both is the same, since they start with \(v_y = 0\) and have the same acceleration \(g\).
    The bullet’s horizontal velocity does not change the fall time (independence of motions).
    They arrive together.
  3. basic

    A ball is kicked at 20 m/s, at 30° above the horizontal. Find the initial \(v_x\) and \(v_y\). Use \(\sin 30^\circ = 0.5\) and \(\cos 30^\circ \approx 0.87\).

    Show solution
    \(v_x = v_0 \cos\theta\) \(= 20 \cdot 0.87 \approx 17.4\,\text{m/s}\)
    \(v_y = v_0\sin\theta = 20 \cdot 0.5 = 10\,\text{m/s}\)
    \(v_x \approx 17.4\,\text{m/s}\); \(v_y = 10\,\text{m/s}\)
  4. basic

    At the highest point of an angled launch, is the ball’s velocity zero?

    Show solution
    No. At the top, only the vertical component vanishes, and we have \(v_y = 0\).
    The horizontal component is still \(v_x = v_0\cos\theta\) (uniform motion), and the acceleration is still \(g\) downward.
    At the top, the velocity is \(v_x \neq 0\), horizontal.
  5. intermediate

    A ball is launched from the ground at 20 m/s at 45°. Find the range and the maximum height. Use \(g = 10\,\text{m/s}^2\) and \(\sin^2 45^\circ = 0.5\).

    Show solution
    \(A = \dfrac{v_0^2\sin(2\theta)}{g}\) \(= \dfrac{400 \cdot \sin 90^\circ}{10}\) \(= \dfrac{400 \cdot 1}{10}\) \(= 40\,\text{m}\)
    \(H = \dfrac{v_0^2\sin^2\theta}{2g}\) \(= \dfrac{400 \cdot 0.5}{20}\) \(= 10\,\text{m}\)
    \(A = 40\,\text{m}\); \(H = 10\,\text{m}\)
  6. intermediate

    A cannon fires at 30 m/s, first at 30° and then at 60°. Compare the ranges and the maximum heights. Use \(g = 10\,\text{m/s}^2\), \(\sin 60^\circ = \sin 120^\circ \approx 0.87\), \(\sin^2 30^\circ = 0.25\) and \(\sin^2 60^\circ = 0.75\).

    Show solution
    Range: \(\sin(2 \cdot 30^\circ) = \sin(2 \cdot 60^\circ) \approx 0.87\), so \(A = \dfrac{900 \cdot 0.87}{10} \approx 78\,\text{m}\) in both cases.
    \(H_{30} = \dfrac{900 \cdot 0.25}{20} = 11.25\,\text{m}\)
    \(H_{60} = \dfrac{900 \cdot 0.75}{20} = 33.75\,\text{m}\)
    Complementary angles: same range (≈ 78 m), but the 60° shot goes 3 times higher.
  7. intermediate

    A rescue plane flies horizontally at 50 m/s, 500 m above the ground, and drops a package. How long does the package take to fall, and how many metres before the target must it be released? Use \(g = 10\,\text{m/s}^2\) and ignore air resistance.

    Show solution
    \(t = \sqrt{\dfrac{2 \cdot 500}{10}} = \sqrt{100} = 10\,\text{s}\)
    \(x = 50 \cdot 10 = 500\,\text{m}\) before the target
    During the fall, the package stays right below the plane, because it has the same \(v_x\).
    10 s; 500 m before the target.
  8. intermediate

    A ball leaves the ground with \(v_x = 12\,\text{m/s}\) and \(v_y = 16\,\text{m/s}\). Find \(v_0\), the flight time, the range and the maximum height. Use \(g = 10\,\text{m/s}^2\).

    Show solution
    \(v_0 = \sqrt{12^2 + 16^2} = \sqrt{400} = 20\,\text{m/s}\)
    Flight time: \(t = \dfrac{2\,v_y}{g} = \dfrac{32}{10} = 3.2\,\text{s}\)
    \(A = 12 \cdot 3.2 = 38.4\,\text{m}\)
    \(H = \dfrac{16^2}{2 \cdot 10} = 12.8\,\text{m}\)
    20 m/s; 3.2 s; 38.4 m; 12.8 m.
  9. challenge

    A ball rolls off a 1.25 m high table and hits the floor 2 m from the base of the table. At what speed did it leave the table? At what speed (magnitude) does it reach the floor? Use \(g = 10\,\text{m/s}^2\).

    Show solution
    \(t = \sqrt{\dfrac{2 \cdot 1.25}{10}} = \sqrt{0.25} = 0.5\,\text{s}\)
    \(v_x = \dfrac{2}{0.5} = 4\,\text{m/s}\)
    On arrival: \(v_y = 10 \cdot 0.5 = 5\,\text{m/s}\)
    \(v = \sqrt{4^2 + 5^2}\) \(= \sqrt{41} \approx 6.4\,\text{m/s}\)
    It left at 4 m/s; it lands at ≈ 6.4 m/s.
  10. challenge

    A long jumper wants to reach 8.1 m. Assuming we can treat the jumper as a particle launched at 45° and landing at the same height, what is the minimum take-off speed? What is the jump’s maximum height? Use \(g = 10\,\text{m/s}^2\) and \(\sin^2 45^\circ = 0.5\).

    Show solution
    At 45°, \(\sin(2\theta) = 1\): \(A = \dfrac{v_0^2}{g}\)
    \(v_0^2 = 8.1 \cdot 10 = 81\) \(\Rightarrow v_0 = 9\,\text{m/s}\)
    \(H = \dfrac{v_0^2\sin^2 45^\circ}{2g}\) \(= \dfrac{81 \cdot 0.5}{20} \approx 2.0\,\text{m}\)
    Since 45° gives the longest range, any other angle would need more speed.
    \(v_0 = 9\,\text{m/s}\); \(H \approx 2\,\text{m}\)
STEP 8

Velocities add like vectors

Since velocity has a magnitude and a direction, it is a vector. When one motion happens ‘inside’ another, such as a boat on flowing water, a plane in the wind or a person on an escalator, the velocity relative to the ground is the vector sum of the two.

Along the same line, the speeds simply add or subtract. A plane at 800 km/h with a 100 km/h tailwind makes 900 km/h, and against a headwind it makes 700 km/h; on an escalator at 0.5 m/s, a person who walks at 1 m/s goes up at 1.5 m/s. In perpendicular directions, the sum follows Pythagoras.

On the river, only the part of the velocity that crosses the river decides the crossing time. The current merely pushes the boat downstream, and we call that shift the drift.

\(\vec v_R = \vec v_b + \vec v_c\)\(v_R = \sqrt{v_b^2 + v_c^2}\) (perpendicular)\(t = \dfrac{L}{v_b\cos\alpha}\)\(\text{drift} = (v_c + v_b\sin\alpha)\,t\)\(\alpha\) is the angle between the bow and the line perpendicular to the banks, negative when the boat points upstream. The river is \(L = 60\,\text{m}\) wide.

boat (relative to the water)   current   resultant

Let's discuss

  • In ‘Point straight across’, check \(v_R\) with Pythagoras and work out the drift in that case.
  • If we increase the current without touching the boat, does the crossing time change?
  • Use ‘Land directly opposite’ and notice where the boat points. Does it arrive faster or slower than when it pointed straight across?
  • With the current stronger than the boat, could there be any angle that takes the boat exactly to the opposite point?
Resultant velocity
Crossing time
Drift
Distance travelled
Exercises for step 8 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A boat crosses a river pointing perpendicular to the bank at 4 m/s, and the current is 3 m/s. What is the boat’s velocity relative to the bank?

    Show solution
    Since the two velocities are perpendicular, we use Pythagoras.
    \(v_R = \sqrt{4^2 + 3^2} = \sqrt{25} = 5\,\text{m/s}\)
    \(v_R = 5\,\text{m/s}\)
  2. basic

    A plane flies at 800 km/h relative to the air. What is its speed relative to the ground with a 100 km/h tailwind? And with a headwind?

    Show solution
    Along the same line, we just add or subtract.
    Tailwind: \(800 + 100 = 900\,\text{km/h}\)
    Headwind: \(800 - 100 = 700\,\text{km/h}\)
    900 km/h with a tailwind; 700 km/h with a headwind.
  3. basic

    A 15 m escalator moves at 0.5 m/s. A person walks up at 1 m/s relative to the steps. How long does it take? And if she stands still on the escalator?

    Show solution
    Walking: \(v = 1 + 0.5 = 1.5\,\text{m/s}\) and \(t = \dfrac{15}{1.5} = 10\,\text{s}\)
    Standing: \(t = \dfrac{15}{0.5} = 30\,\text{s}\)
    10 s walking; 30 s standing.
  4. basic

    A boat crosses a river always pointing perpendicular to the banks. If the current gets stronger, does the crossing time change? What changes?

    Show solution
    The time depends only on the velocity component that crosses the river, which is the boat’s: \(t = \dfrac{L}{v_b}\).
    The current is parallel to the banks and, for that reason, neither helps nor hinders the crossing.
    What increases is the drift: the boat lands farther downstream.
    The time does not change; the drift increases.
  5. intermediate

    A river is 60 m wide with a 2 m/s current. A boat points perpendicular to the banks at 3 m/s. Find the crossing time, the drift and the total distance travelled relative to the banks.

    Show solution
    \(t = \dfrac{60}{3} = 20\,\text{s}\)
    Drift: \(2 \cdot 20 = 40\,\text{m}\) downstream
    \(d = \sqrt{60^2 + 40^2}\) \(= \sqrt{5200} \approx 72.1\,\text{m}\)
    20 s; 40 m of drift; ≈ 72.1 m.
  6. intermediate

    A boat at 5 m/s (relative to the water) wants to cross an 80 m river and land exactly opposite, with a 3 m/s current. Where should it point? What is its velocity relative to the bank, and how long does it take?

    Show solution
    It must point a little upstream, so that this component cancels the current:
    \(5\sin\alpha = 3\) \(\Rightarrow \sin\alpha = 0.6\) \(\Rightarrow \alpha \approx 37^\circ\) from the perpendicular
    What remains is the crossing component: \(v_R = \sqrt{5^2 - 3^2} = 4\,\text{m/s}\)
    \(t = \dfrac{80}{4} = 20\,\text{s}\)
    Point ≈ 37° upstream; 4 m/s; 20 s.
  7. intermediate

    A plane flies north at 400 km/h relative to the air, and the wind blows east at 300 km/h. What is the plane’s velocity relative to the ground? How far does it travel in 2 h?

    Show solution
    \(v_R = \sqrt{400^2 + 300^2}\) \(= \sqrt{250\,000}\) \(= 500\,\text{km/h}\)
    Deviation: \(\tan\theta = \dfrac{300}{400} = 0.75\) \(\Rightarrow \theta \approx 37^\circ\) east of north
    \(d = 500 \cdot 2 = 1000\,\text{km}\)
    500 km/h; 1000 km in 2 h.
  8. intermediate

    Rain falls vertically at 8 m/s. A car moves at 6 m/s. For the driver, at what speed and at what angle does the rain fall?

    Show solution
    Relative to the car, the rain gains a horizontal component of 6 m/s, coming from the front.
    \(v = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\,\text{m/s}\)
    \(\tan\theta = \dfrac{6}{8} = 0.75\) \(\Rightarrow \theta \approx 37^\circ\) from the vertical
    That is why raindrops streak diagonally across the side window.
    10 m/s, tilted ≈ 37° from the vertical.
  9. challenge

    Standing still on an escalator, a person goes up in 30 s. Walking with the escalator switched off, she goes up in 20 s. How long does it take walking with the escalator switched on?

    Show solution
    Calling the length \(L\), \(v_{\text{esc}} = \dfrac{L}{30}\) and \(v_{\text{walk}} = \dfrac{L}{20}\)
    \(v = \dfrac{L}{30} + \dfrac{L}{20} = \dfrac{2L + 3L}{60} = \dfrac{L}{12}\)
    \(t = 12\,\text{s}\)
  10. challenge

    A 100 m wide river has a 2 m/s current; a boat moves at 4 m/s relative to the water. (a) Pointing perpendicular to the banks, how long does it take and what is the drift? (b) To land exactly opposite, how long does it take? Which strategy is faster?

    Show solution
    (a) \(t = \dfrac{100}{4} = 25\,\text{s}\); drift \(= 2 \cdot 25 = 50\,\text{m}\)
    (b) \(v_R = \sqrt{4^2 - 2^2}\) \(= \sqrt{12} \approx 3.46\,\text{m/s}\); \(t = \dfrac{100}{3.46} \approx 28.9\,\text{s}\)
    Pointing perpendicular is faster (25 s); landing directly opposite costs more time (≈ 28.9 s).
WRAP-UP

Challenges

Reference frame and average velocity

A bus goes from km 10 to km 70 in 45 min and then returns to km 40 in 30 min. Find the displacement, the distance travelled, the average speed and the average velocity.

Show solution
\(\Delta s = 40 - 10 = 30\,\text{km}\)
\(d = 60 + 30 = 90\,\text{km}\)
\(\Delta t = 45 + 30 = 75\,\text{min} = 1.25\,\text{h}\)
Average speed: \(\dfrac{90}{1.25} = 72\,\text{km/h}\)
Average velocity: \(\dfrac{30}{1.25} = 24\,\text{km/h}\)
Uniform + accelerated motion

A car is moving at 108 km/h when the driver looks at a phone for 2 s and only then brakes, at \(6\,\text{m/s}^2\). How many metres does the car travel from the start of the distraction until it stops?

Show solution
\(108\,\text{km/h} = 30\,\text{m/s}\)
Distraction (uniform motion): \(30 \cdot 2 = 60\,\text{m}\)
Braking (Torricelli): \(\Delta s = \dfrac{30^2}{2 \cdot 6} = 75\,\text{m}\)
Total: \(60 + 75 = 135\,\text{m}\)
Meeting with acceleration

A parked police car starts with \(a = 4\,\text{m/s}^2\) at the instant a car passes it at a constant 20 m/s. When and where does the police car catch the car, and at what speed?

Show solution
\(s_V = \dfrac{4}{2}\,t^2 = 2t^2\) and \(s_C = 20t\)
\(2t^2 = 20t\) \(\Rightarrow t = 10\,\text{s}\)
\(s = 20 \cdot 10 = 200\,\text{m}\)
\(v_V = 4 \cdot 10 = 40\,\text{m/s} = 144\,\text{km/h}\)
Free fall

A raindrop falls from a cloud 2000 m high. If there were no air, at what speed would it reach the ground, and why does that not happen? Use \(g = 10\,\text{m/s}^2\).

Show solution
\(v = \sqrt{2gh}\) \(= \sqrt{2 \cdot 10 \cdot 2000}\) \(= \sqrt{40\,000}\) \(= 200\,\text{m/s}\)
\(200 \cdot 3.6 = 720\,\text{km/h}\): it would be like a bullet!
The air slows the drop, which soon reaches a terminal velocity of just a few metres per second.
Horizontal launch

A ball rolls off the top of a 45 m cliff horizontally at 10 m/s. How long does it take to fall, how far from the base does it land, and at what speed? Use \(g = 10\,\text{m/s}^2\).

Show solution
\(t = \sqrt{\dfrac{2 \cdot 45}{10}} = 3\,\text{s}\)
\(x = 10 \cdot 3 = 30\,\text{m}\)
\(v_y = 10 \cdot 3 = 30\,\text{m/s}\)
\(v = \sqrt{10^2 + 30^2}\) \(= \sqrt{1000} \approx 31.6\,\text{m/s}\)
Vectors

A boat crosses a 120 m river pointing perpendicular to the banks at 4 m/s, in a current of 3 m/s. Find the time, the drift, the velocity relative to the bank and the distance travelled.

Show solution
\(t = \dfrac{120}{4} = 30\,\text{s}\)
Drift: \(3 \cdot 30 = 90\,\text{m}\)
\(v_R = \sqrt{4^2 + 3^2} = 5\,\text{m/s}\)
\(d = 5 \cdot 30 = 150\,\text{m}\) (check: \(\sqrt{120^2 + 90^2} = 150\))
v × t graph

A car goes from rest to 20 m/s in 5 s, holds 20 m/s for 10 s and brakes to a stop in 4 s. Find the acceleration on each leg, the total displacement and the average velocity.

Show solution
Accelerations: \(\dfrac{20}{5} = 4\,\text{m/s}^2\); \(0\); \(-\dfrac{20}{4} = -5\,\text{m/s}^2\)
Areas: \(\dfrac{5 \cdot 20}{2} = 50\,\text{m}\); \(10 \cdot 20 = 200\,\text{m}\); \(\dfrac{4 \cdot 20}{2} = 40\,\text{m}\)
\(\Delta s = 290\,\text{m}\) in \(19\,\text{s}\)
\(v_m = \dfrac{290}{19} \approx 15.3\,\text{m/s}\)
Think, no maths

Inside a bus moving in a straight line at constant speed, you toss a coin straight up. Does it land in your hand, behind it or in front of it, and how does a person on the pavement see its path? It may help to picture the scene before doing any maths.

Show solution
It lands in your hand, because the coin already has the same horizontal velocity as you and the bus, and keeps that velocity in the air (independence of motions).
For you, it goes up and down in a straight line.
For someone on the pavement, it traces an arc of a parabola, moving along with the bus.
If the bus brakes during the flight, then the coin does land in front of you.

ENEM-style questions

The ENEM is Brazil’s national secondary-school exam, the one most students sit to get into university, and we wrote these five questions in its format, with a short everyday text, a question and five options of which only one is right. The wrong options repeat mistakes we see in class, so it is worth reading the solution even when you get the answer. Further down there are real questions from the exam to practise with.

  1. Average velocity · Step 2

    Running apps use the phone’s GPS to record the distance and time of each run and, at the end, show an average speed. On a 10 km run, Camila covered the first 5 km in 25 min, stopped for 5 min at a water fountain and ran the last 5 km in 30 min, without pausing the app’s timer.

    If the app works out the average speed of the run the way we do in physics, what value, in km/h, should it show?

    1. 2.8
    2. 10.0
    3. 10.9
    4. 11.0
    5. 12.0
    Show solution
    Answer: B.
    The average always uses the total time, and the stop at the fountain counts.
    \(\Delta t = 25 + 5 + 30 = 60\,\text{min} = 1\,\text{h}\)
    \(v_m = \dfrac{\Delta s}{\Delta t}\) \(= \dfrac{10\,\text{km}}{1\,\text{h}}\) \(= 10.0\,\text{km/h}\)
    Choosing C means leaving out the stop \(\left(\dfrac{10}{55/60} \approx 10.9\right)\), D takes the plain average of 12 and 10 km/h, E keeps only the first leg, and A gives the value in m/s \((10/3.6 \approx 2.8)\) as if it were km/h. Anyone who averaged the two legs is in for a surprise.
  2. When two objects meet · Step 4

    Many avenues share their space between a bus lane and a cycle lane, and GPS can track both vehicles at once. On a straight avenue, a set of traffic lights serves as the origin, and the GPS records every 10 s the position of a bus and that of Larissa, a cyclist, both moving in the same direction.

    Positions recorded by GPS
    t (s)0102030
    bus (m)080160240
    Larissa (m)300350400450

    If both keep the speeds shown in the table, at what instant and at what position does the bus catch up with Larissa?

    1. After about 23 s, near position 185 m.
    2. After 37.5 s, at position 300 m.
    3. After 60 s, at position 480 m.
    4. After 100 s, at position 500 m.
    5. After 100 s, at position 800 m.
    Show solution
    Answer: E.
    The positions change by the same amount every 10 s. Both are therefore in uniform motion, and we can read the speeds off the table.
    \(v_{\text{bus}} = \dfrac{80}{10} = 8\,\text{m/s}\) and \(v_L = \dfrac{50}{10} = 5\,\text{m/s}\)
    \(s_{\text{bus}} = 8t\) and \(s_L = 300 + 5t\)
    \(8t = 300 + 5t\) \(\Rightarrow 3t = 300\) \(\Rightarrow t = 100\,\text{s}\)
    \(s = 8 \cdot 100 = 800\,\text{m}\)
    A adds the speeds, as if the two were heading towards each other; B forgets that Larissa is moving too; C divides the 300 m by her speed; D gets the time right and works out the position without the \(s_0 = 300\,\text{m}\).
  3. Acceleration and the v × t graph · Step 5

    Between two stations, an underground train starts from rest, speeds up, runs for a while at constant speed and brakes until it stops at the next platform. The graph shows, in simplified form, the speed that the control system recorded along this journey.

    0 20 60 70 20 0 t (s) v (m/s)

    Based on the graph, what is the train’s acceleration as it pulls away, and how far apart are the two stations?

    1. 1 m/s² and 1100 m.
    2. 1 m/s² and 1400 m.
    3. 20 m/s² and 1100 m.
    4. 2 m/s² and 1100 m.
    5. 1 m/s² and 800 m.
    Show solution
    Answer: A.
    The acceleration as it pulls away is the slope of the first part of the graph.
    \(a = \dfrac{\Delta v}{\Delta t} = \dfrac{20 - 0}{20} = 1\,\text{m/s}^2\)
    The distance is the area under the graph, which we can split into two triangles and a rectangle.
    \(\Delta s = \dfrac{20 \cdot 20}{2} + 40 \cdot 20 + \dfrac{10 \cdot 20}{2}\)
    \(\Delta s = 200 + 800 + 100 = 1100\,\text{m}\)
    B treats the area as a single rectangle \((20 \cdot 70)\), C mixes up velocity and acceleration, D uses the slope of the braking part \((20/10 = 2)\) and E keeps only the constant-speed part.
  4. Reaction time and braking · Steps 3 and 5

    The distance a car covers before it stops has two parts. While the driver reacts, the foot has not yet reached the brake and the car carries on at the same speed; after that, the brakes produce a deceleration that, on dry tarmac, we can assume is constant. Diego is driving at 72 km/h. When he is paying attention, he takes 0.8 s to react. Let’s assume that, while talking on the phone, this time rises to 2.0 s, and that in both cases the brakes slow the car at 8 m/s².

    Distracted by the phone, how many metres does Diego travel from the instant a pedestrian appears ahead until the car stops?

    1. 25
    2. 40
    3. 41
    4. 65
    5. 90
    Show solution
    Answer: D.
    In SI units, \(72\,\text{km/h} = 20\,\text{m/s}\).
    During the reaction time the car is in uniform motion.
    \(d_r = 20 \cdot 2.0 = 40\,\text{m}\)
    The braking distance comes from Torricelli’s equation, with a final speed of zero.
    \(0 = 20^2 - 2 \cdot 8 \cdot d_f\) \(\Rightarrow d_f = \dfrac{400}{16} = 25\,\text{m}\)
    \(d = 40 + 25 = 65\,\text{m}\)
    A and B keep only one of the two parts, C is the distance for an attentive Diego \((16 + 25)\) and E comes from forgetting the 2 in Torricelli’s equation \((400/8 = 50)\). The phone adds 24 m, close to the length of five cars.
  5. Free fall and projectiles · Steps 6 and 7

    Drones are already being tested to carry medicines and first-aid kits to places that are hard to reach. In one test, a drone flies horizontally at 12 m/s, 20 m above the ground, and drops a compact parcel. We will ignore air resistance, which seems reasonable for a small, heavy parcel, and use g = 10 m/s².

    How far, horizontally, from the point where it was released does the parcel hit the ground?

    1. 0 m
    2. 17 m
    3. 24 m
    4. 40 m
    5. 48 m
    Show solution
    Answer: C.
    Once released, the parcel keeps the drone’s horizontal velocity, and the motion splits in two. Vertically, it is a free fall from rest.
    \(t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 \cdot 20}{10}} = 2\,\text{s}\)
    Horizontally, the parcel moves uniformly.
    \(x = v_x\,t = 12 \cdot 2 = 24\,\text{m}\)
    A assumes the parcel falls straight down, which, without air, never happens to something dropped from a moving drone; B forgets the 2 inside the square root \((t \approx 1.4\,\text{s})\), D uses the vertical speed on landing \((\sqrt{2gh} = 20\,\text{m/s})\) instead of \(v_x\) and E forgets the square root \((t = 4\,\text{s})\). Two seconds of falling may seem short.

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, where each question can be found by year, day, booklet colour and number. Until 2016, the Natural Sciences paper was sat on Day 1.

  • ENEM 2012, Day 1, blue booklet, question 60. An underground train speeds up, holds its speed and brakes between two stations, and the question asks for the \(s \times t\) graph that matches this motion.
  • ENEM 2012, Day 1, blue booklet, question 72. A delivery runs along two stretches with different speed limits, and the question asks for the total journey time.
  • ENEM 2016, Day 1, blue booklet, question 63. A car stopping in two stages, reaction and braking, becomes a graph of speed against distance, and the task is to pick the correct sketch.
  • ENEM 2017, Day 2, blue booklet, question 131. The question compares the stopping distance of an attentive driver with that of one who is using a phone and takes longer to start braking.
  • ENEM 2022, Day 2, blue booklet, question 100. The range of a water jet fired horizontally is used to estimate how high the same jet would reach if pointed straight up.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Displacement\(\Delta s = s - s_0\)
Average velocity\(v_m = \Delta s / \Delta t\)
Units\(1\,\text{m/s} = 3.6\,\text{km/h}\)
Uniform motion\(s = s_0 + v\,t\)
Meeting\(s_A = s_B\)
Acceleration\(a = \Delta v / \Delta t\)
Accelerated motion: velocity\(v = v_0 + a\,t\)
Accelerated motion: position\(s = s_0 + v_0 t + \frac{a t^2}{2}\)
Torricelli (the v² equation)\(v^2 = v_0^2 + 2a\,\Delta s\)
Free fall (g ≈ 10 m/s²)\(t_q = \sqrt{2h/g}\quad H = \frac{v_0^2}{2g}\)
Angled launch\(v_x = v_0\cos\theta\quad A = \frac{v_0^2\sin 2\theta}{g}\)
Adding (perpendicular)\(v_R = \sqrt{v_1^2 + v_2^2}\)