Senior 2 practice book · Unit 2 of 10

Uniformly Accelerated Rectilinear Motion

50 questions that complete the Senior 2 quiz for this unit: 21 core and 29 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

If a body's velocity is zero, its acceleration must also be zero.At the top of its flight a ball has zero velocity but its acceleration is still g downwards; that is why it starts to fall back.
A negative acceleration always means the body is slowing down.The sign only gives the direction. If the velocity is also negative (moving backwards), a negative acceleration makes the body speed up.
A steeper line on any graph means a faster body.Only on a displacement–time graph does the gradient give velocity. On a velocity–time graph a steeper line means a larger acceleration.
Average speed is the average of the speeds.Average speed = total distance ÷ total time. If you spend longer at the slow speed, the average is closer to the slow speed.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Distance and displacement; speed and velocity
Acceleration and deceleration: meaning and SI unit
Equations of uniformly accelerated motion and choosing the right one
Velocity–time graphs: gradient gives acceleration, area gives displacement
Displacement–time graphs of accelerated motion: curves and tangents
Acceleration–time graphs
Free fall: acceleration due to gravity and bodies dropped from rest
Bodies thrown vertically upwards: time to the top, maximum height, return
Ticker-timer tapes: measuring velocity and acceleration in the laboratory
Stopping distance: thinking distance and braking distance
Average speed and average velocity for journeys with several stages
Signs and directions: positive and negative velocity and acceleration
Converting speeds between km/h and m/s

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 29 harder ones are marked ★ Challenge.

1True or false
If the dots on a ticker tape are equally spaced, the trolley moved with uniform velocity.
Show answer
Answer: True
Equal distances in equal times (0.02 s) in one direction means uniform velocity.
Common mistakeLearners sometimes think equal spacing means uniform acceleration; uniform acceleration gives spacing that increases by equal amounts.
2True or false · ★ Challenge
A body with a negative acceleration must be slowing down.
Show answer
Answer: False
If the velocity is also negative (moving the other way), a negative acceleration makes it speed up in that direction.
Common mistakeLearners treat 'negative acceleration' as another name for deceleration; deceleration means acceleration opposite to the velocity.
3True or false · ★ Challenge
Two velocity–time lines that are parallel but start at different velocities show the same acceleration.
Show answer
Answer: True
Parallel lines have the same gradient, and the gradient of a v–t graph is the acceleration.
Common mistakeLearners often think the higher line has more acceleration; height shows velocity, slope shows acceleration.
4True or false
Two cars moving at 60 km/h in opposite directions on the same road have the same velocity.
Show answer
Answer: False
They have the same speed but opposite directions, so their velocities are different (+60 km/h and −60 km/h).
Common mistakeLearners who ignore direction treat equal speeds as equal velocities.
5True or false · ★ Challenge
For a round trip that ends where it started, the average velocity is zero even though the average speed is not.
Show answer
Answer: True
Displacement is zero at the end, so average velocity = 0 ÷ time = 0, while the distance travelled is not zero.
Common mistakeLearners mix up average speed (distance ÷ time) and average velocity (displacement ÷ time).
6Fill in the blank
A ticker-timer connected to a 50 Hz supply makes ______ dots every second.
Show answer
Answer: 50
The vibrator strikes the tape once per cycle, 50 times each second.
Common mistakeWriting 0.02 gives the time between dots, not the number of dots per second.
7Multiple choice · ★ Challenge
Which of these has the greatest acceleration?
1. AA bicycle: 0 to 6 m/s in 3 s
2. BA plane: 0 to 75 m/s in 30 s
3. CA car: 0 to 27 m/s in 9 s
4. DA runner: 0 to 9 m/s in 4.5 s
Show answer
Answer: C. A car: 0 to 27 m/s in 9 s
a = Δv ÷ t: car 3 m/s², bicycle 2 m/s², plane 2.5 m/s², runner 2 m/s².
Common mistakeChoosing the plane because it reaches the highest speed ignores the time; acceleration is the RATE of change of velocity.
8Fill in the blank · ★ Challenge
A body falls from rest without air resistance. The distances it falls in the 1st, 2nd, 3rd and 4th seconds are in the ratio 1 : 3 : 5 : ______.
Show answer
Answer: 7
Total distances after 1, 2, 3, 4 s are ½g × 1, 4, 9, 16; the distances in each second are 1, 3, 5, 7 times ½g.
Common mistakeWriting 16 gives the total distance after 4 s (ratio 1 : 4 : 9 : 16), not the distance in the 4th second alone.
9Fill in the blank
To change a speed from km/h to m/s, divide it by ______.
Show answer
Answer: 3.6
1 km/h = 1000 m ÷ 3600 s = 1/3.6 m/s.
Common mistakeDividing by 60 or by 1000 alone converts only the time or only the distance.
10Multiple choice · ★ Challenge
A body starts from rest and accelerates uniformly. After travelling a distance s its speed is v. What is its speed after travelling 4s from the start?
1. A4v
2. B2v
3. C16v
4. D8v
Show answer
Answer: B. 2v
v² = 2as, so v is proportional to √s. Four times the distance gives √4 = 2 times the speed.
Common mistakeChoosing 4v assumes speed is proportional to distance; for uniform acceleration from rest it is proportional to the square root of distance.
11Multiple choice
Which factor increases a driver's thinking distance but NOT the braking distance?
1. AThe driver is tired or has drunk alcohol
2. BThe road surface is wet
3. CThe tyres are worn smooth
4. DThe vehicle carries a heavy load
Show answer
Answer: A. The driver is tired or has drunk alcohol
Tiredness and alcohol slow the driver's reaction, so the car travels further before the brakes act; the other factors affect the braking itself.
Common mistakeChoosing the wet road is tempting because it is dangerous, but it affects braking (less grip), not reaction time.
12Multiple choice · ★ Challenge
A driver moving at 15 m/s has a reaction time of 0.8 s. What is her thinking distance?
1. A18.8 m
2. B15 m
3. C12 m
4. D9.4 m
Show answer
Answer: C. 12 m
During the reaction time the speed is constant: s = vt = 15 × 0.8 = 12 m.
Common mistakeChoosing 18.8 m divides 15 by 0.8 instead of multiplying; distance = speed × time.
13Multiple choice · ★ Challenge
A ball is thrown straight up and caught at the same height 3 s later (g = 10 m/s², no air resistance). At what speed was it thrown?
1. A30 m/s
2. B10 m/s
3. C15 m/s
4. D45 m/s
Show answer
Answer: C. 15 m/s
Up and down take equal times, so it rises for 1.5 s; u = gt = 10 × 1.5 = 15 m/s.
Common mistakeChoosing 30 m/s uses the whole 3 s for the upward journey; the ball only rises for half the flight time.
14Multiple choice
Starting from the first dot made, the dots on a ticker tape get further and further apart. The trolley was:
1. AMoving at steady speed
2. BSlowing down
3. CSpeeding up
4. DStationary
Show answer
Answer: C. Speeding up
Each space takes the same time, so bigger spaces mean more distance per 0.02 s: the trolley is speeding up.
Common mistakeSome learners read the tape backwards; check which end was pulled through the timer first.
15Multiple choice · ★ Challenge
A stone dropped down a well hits the water 1.5 s later (g = 10 m/s², ignore air resistance). How deep is the water surface below the top?
1. A15 m
2. B22.5 m
3. C7.5 m
4. D11.25 m
Show answer
Answer: D. 11.25 m
h = ½gt² = ½ × 10 × 1.5² = 5 × 2.25 = 11.25 m.
Common mistakeChoosing 22.5 m forgets the ½ in h = ½gt²; 15 m is the final speed gt, not a distance.
16Fill in the blank
A car moving forwards has an acceleration of −3 m/s². This is called a ______ of 3 m/s².
Show answer
Answer: deceleration (retardation)
An acceleration opposite to the velocity slows the car: a deceleration.
Common mistakeWriting 'negative velocity' confuses the sign of the acceleration with the direction of motion.
17Short answer · ★ Challenge
A flat sheet of paper falls slowly, but the same sheet crumpled into a ball falls much faster, although its mass is unchanged. Explain.
Show answer
Model answer: Both have the same weight, but the flat sheet has a large area facing the air, so air resistance on it is large and soon balances its weight at a low speed. The crumpled ball has a small area, so air resistance is small and it falls with an acceleration close to g.
Common mistakeSaying the ball is 'heavier' is wrong: the mass is the same; only the shape and the air resistance have changed.
18Multiple choice
On the Moon there is no air and g is about 1.6 m/s². A hammer and a feather are dropped together from the same height. They:
1. ALand with the hammer first, as it is heavier
2. BLand together, taking less time than on Earth
3. CLand together, taking longer than on Earth
4. DDo not fall, as the Moon has no air
Show answer
Answer: C. Land together, taking longer than on Earth
Without air resistance all bodies fall with the same acceleration; with g smaller than on Earth, the fall takes longer.
Common mistakeChoosing 'do not fall' confuses air with gravity; the Moon still pulls objects down, it just has no air to slow the feather.
19Multiple choice · ★ Challenge
Taking upwards as positive, a ball is thrown up at 12 m/s (g = 10 m/s²). What is its velocity 2 s later?
1. A+8 m/s, moving up
2. B+32 m/s, moving up
3. C0 m/s, at the top
4. D−8 m/s, moving down
Show answer
Answer: D. −8 m/s, moving down
v = u + at = 12 + (−10) × 2 = 12 − 20 = −8 m/s; the minus sign means it is moving down.
Common mistakeChoosing +32 m/s takes g as +10 m/s²; with upwards positive, g is −10 m/s².
20Multiple choice
A cheetah can sprint at 25 m/s. Expressed in kilometres per hour, this is:
1. A6.9 km/h
2. B250 km/h
3. C1500 km/h
4. D90 km/h
Show answer
Answer: D. 90 km/h
25 m/s × 3.6 = 90 km/h.
Common mistakeChoosing 6.9 km/h divides by 3.6; to go from m/s to km/h you MULTIPLY by 3.6.
21Multiple choice · ★ Challenge
On a tape from a 50 Hz ticker-timer, a section of 10 dot-spaces is 8.0 cm long. What was the average velocity during this section?
1. A0.40 m/s
2. B0.80 m/s
3. C0.16 m/s
4. D4.0 m/s
Show answer
Answer: A. 0.40 m/s
Time = 10 × 0.02 = 0.20 s; v = 0.080 m ÷ 0.20 s = 0.40 m/s.
Common mistakeChoosing 0.80 m/s uses 0.1 s (five spaces) instead of 0.2 s; 4.0 m/s takes 8.0 cm as 0.80 m instead of 0.080 m.
22Multiple choice · ★ Challenge
A cyclist rides 6 km uphill at 12 km/h, then returns the same 6 km downhill at 36 km/h. What is her average speed for the round trip?
1. A18 km/h
2. B24 km/h
3. C0 km/h
4. D21 km/h
Show answer
Answer: A. 18 km/h
Times: 6 ÷ 12 = 0.5 h and 6 ÷ 36 = 1/6 h; total 2/3 h. Average speed = 12 km ÷ 2/3 h = 18 km/h.
Common mistakeChoosing 24 km/h averages the two speeds; she spends three times longer going uphill, so the average is nearer 12 km/h.
23Multiple choice
An athlete runs one and a half laps of a 400 m circular track. Which statement is correct?
1. ADistance 600 m; displacement also equal to 600 m
2. BDistance 600 m; displacement equal to the track's diameter
3. CDistance 600 m; displacement zero, as the track is round
4. DDistance 400 m; displacement 200 m along the track
Show answer
Answer: B. Distance 600 m; displacement equal to the track's diameter
Distance is the path length, 1.5 × 400 = 600 m. After half a lap extra she is on the opposite side, so her displacement is straight across: one diameter.
Common mistakeChoosing 'displacement zero' would be true after whole laps only; after 1.5 laps she is not back at the start.
24Multiple choice · ★ Challenge
Already cruising at 15 m/s on the Kigali–Rubavu road, a minibus driver presses the accelerator to overtake, gaining speed at 1.5 m/s² for 8 s. What is the minibus's displacement during the overtaking?
1. A168 m
2. B48 m
3. C216 m
4. D120 m
Show answer
Answer: A. 168 m
s = ut + ½at² = 15 × 8 + ½ × 1.5 × 8² = 120 + 48 = 168 m.
Common mistakeChoosing 48 m takes u = 0; the minibus was already moving at 15 m/s, which adds ut = 120 m.
25Multiple choice
Which equation gives the time taken when u, v and a are known?
1. At = (v + u) ÷ a
2. Bt = a ÷ (v − u)
3. Ct = 2s ÷ (u + v)
4. Dt = (v − u) ÷ a
Show answer
Answer: D. t = (v − u) ÷ a
From v = u + at, at = v − u, so t = (v − u) ÷ a.
Common mistaket = 2s ÷ (u + v) is a correct equation, but it needs s, which is not known here.
26Multiple choice · ★ Challenge
Part of a velocity–time graph lies below the time axis. What does this show?
1. AThe body is moving in the opposite direction
2. BThe body is slowing down
3. CThe body is not moving at all
4. DThe body has zero acceleration
Show answer
Answer: A. The body is moving in the opposite direction
Negative values of velocity mean motion in the negative (opposite) direction.
Common mistakeChoosing 'slowing down' is tempting, but a body below the axis can be speeding up backwards; slowing down shows as the line moving TOWARDS the axis.
27Fill in the blank
For a body that starts from rest (u = 0), the equation v² = u² + 2as becomes v² = ______.
Show answer
Answer: 2as
With u = 0, the u² term is zero, leaving v² = 2as.
Common mistakeWriting v = 2as forgets that the left side is v squared.
28Multiple choice · ★ Challenge
The area under an acceleration–time graph between two times gives:
1. AThe displacement
2. BThe change in velocity
3. CThe distance travelled
4. DThe rate of change of acceleration
Show answer
Answer: B. The change in velocity
Area = a × t, and a × t = Δv, the change in velocity.
Common mistakeChoosing displacement is tempting because the area under a v–t graph gives displacement; under an a–t graph it gives Δv.
29Multiple choice · ★ Challenge
A displacement–time graph curves over and becomes horizontal at t = 6 s. What is happening to the body at 6 s?
1. AIt has come to rest
2. BIt has its largest acceleration
3. CIt has its greatest speed
4. DIt is back at its starting point
Show answer
Answer: A. It has come to rest
A horizontal line has zero gradient, so the velocity at 6 s is zero: the body has stopped.
Common mistakeChoosing 'back at its starting point' confuses zero gradient with zero displacement; the line is horizontal ABOVE the axis.
30Multiple choice
Taking upwards as positive, what is the acceleration of a stone thrown upwards while it is still rising? (g = 10 m/s²)
1. A+10 m/s²
2. B0 m/s²
3. C−10 m/s²
4. D−20 m/s²
Show answer
Answer: C. −10 m/s²
Gravity pulls downwards all the time, so a = −10 m/s² whether the stone is rising or falling.
Common mistakeChoosing +10 m/s² because the stone moves upwards confuses the direction of motion with the direction of the acceleration.
31Short answer · ★ Challenge
Explain how you would find the velocity of a body at one particular instant from a curved displacement–time graph.
Show answer
Model answer: Draw a tangent to the curve at that instant (a straight line that just touches it). Choose two points far apart on the tangent and find its gradient Δs ÷ Δt; this is the velocity at that instant.
Common mistakeDividing the displacement at that point by the time (s ÷ t) gives the average velocity from the start, not the velocity at that instant.
32Multiple choice
A ticker-timer makes 50 dots every second. What is the time interval between two neighbouring dots?
1. A0.5 s
2. B50 s
3. C0.2 s
4. D0.02 s
Show answer
Answer: D. 0.02 s
Time between dots = 1 ÷ 50 = 0.02 s.
Common mistakeChoosing 0.2 s confuses one dot-space with a 10-space section (10 × 0.02 = 0.2 s).
33Multiple choice · ★ Challenge
A tangent drawn to a displacement–time curve at t = 4 s passes through the points (2 s, 6 m) and (6 s, 26 m). What is the velocity at t = 4 s?
1. A6.5 m/s
2. B4.3 m/s
3. C20 m/s
4. D5 m/s
Show answer
Answer: D. 5 m/s
Gradient of tangent = (26 − 6) ÷ (6 − 2) = 20 ÷ 4 = 5 m/s.
Common mistakeChoosing 6.5 m/s divides 26 m by 4 s, mixing a point with a time; the gradient needs the changes Δs and Δt.
34Fill in the blank · ★ Challenge
A trolley has a constant acceleration of 3 m/s² for 4 s. The area under its acceleration–time graph is 12 m/s, which is its change in ______.
Show answer
Answer: velocity
Area = 3 m/s² × 4 s = 12 m/s = Δv.
Common mistakeWriting 'distance' is wrong: the unit m/s shows the area is a velocity.
35Multiple choice
A moto's speed increases by 2 m/s every second. What is its acceleration?
1. A2 m/s²
2. B2 m/s
3. C4 m/s²
4. D0.5 m/s²
Show answer
Answer: A. 2 m/s²
Acceleration = change in velocity per second = 2 m/s per second = 2 m/s².
Common mistakeChoosing 2 m/s gives the right number with a velocity unit; acceleration is measured in m/s².
36Multiple choice · ★ Challenge
A train starts from rest, accelerates at 0.5 m/s² for 40 s, then moves at the speed it has reached for 2 minutes. How far does it travel altogether?
1. A2400 m
2. B2800 m
3. C3200 m
4. D400 m
Show answer
Answer: B. 2800 m
v = 0.5 × 40 = 20 m/s; s₁ = ½ × 0.5 × 40² = 400 m; s₂ = 20 × 120 = 2400 m; total = 2800 m.
Common mistakeChoosing 2400 m forgets the accelerating stage; 3200 m takes the first stage as 20 × 40 = 800 m, as if the train had full speed from the start.
37True or false
A horizontal line on a displacement–time graph means that the body is not moving.
Show answer
Answer: True
The displacement does not change with time, so the velocity (gradient) is zero.
Common mistakeLearners who read it like a velocity–time graph think a horizontal line means 'steady speed'.
38Multiple choice · ★ Challenge
On a velocity–time graph, the velocity rises in a straight line from 4 m/s at t = 0 to 14 m/s at t = 10 s. What is the displacement in this time?
1. A140 m
2. B70 m
3. C50 m
4. D90 m
Show answer
Answer: D. 90 m
Area of trapezium = ½(4 + 14) × 10 = 90 m.
Common mistakeChoosing 70 m uses only the triangle ½ × 14 × 10 and forgets that the body started at 4 m/s, not from rest.
39Short answer · ★ Challenge
A girl on a balcony tosses a ball vertically upwards at 25 m/s. Her friend leans out of a window 20 m above the balcony (g = 10 m/s², no air resistance). At what times after the throw is the ball level with the friend? Explain why there are two answers.
Show answer
Model answer: s = ut − ½gt²: 20 = 25t − 5t², so t² − 5t + 4 = 0, (t − 1)(t − 4) = 0, t = 1 s and t = 4 s. It passes the window once on the way up (1 s) and again on the way down (4 s).
Common mistakeKeeping only one answer misses that the ball passes the same height twice; both roots of the equation are real times.
40Short answer
A moto taxi driver says, 'My speedometer shows my velocity.' Explain why this is not quite correct.
Show answer
Model answer: A speedometer shows only the size of the motion (speed). Velocity also needs a direction, which the speedometer does not give; two motos at 40 km/h going opposite ways have the same speed but different velocities.
Common mistakeTreating speed and velocity as the same word ignores that velocity is a vector with direction.
41Short answer · ★ Challenge
Taking forwards as positive, a bus reverses out of a parking place, speeding up from rest to 2 m/s in 4 s. State the signs of its velocity and of its acceleration, and calculate the acceleration.
Show answer
Model answer: Velocity is negative (it moves backwards): v = −2 m/s. a = (v − u) ÷ t = (−2 − 0) ÷ 4 = −0.5 m/s². The acceleration is negative although the bus is speeding up, because velocity and acceleration point the same (backward) way.
Common mistakeGiving +0.5 m/s² because 'the bus speeds up' ignores the direction; the sign shows direction, not speeding up or slowing down.
42Multiple choice
What does the acceleration–time graph look like for a body moving with uniform acceleration?
1. AA straight line sloping up from the origin
2. BA horizontal straight line
3. CA curve that gets steeper
4. DA vertical line at t = 0
Show answer
Answer: B. A horizontal straight line
Uniform acceleration means a does not change, so a plotted against t is a horizontal line.
Common mistakeChoosing the sloping line mixes up the a–t graph with the v–t graph, where uniform acceleration gives a straight sloping line.
43Short answer · ★ Challenge
A lorry on the Kigali–Huye road slows uniformly from 25 m/s to 15 m/s over a distance of 200 m. Find its deceleration and the time taken.
Show answer
Model answer: v² = u² + 2as: 15² = 25² + 2a × 200, so 225 = 625 + 400a, a = −1.0 m/s² (deceleration 1.0 m/s²). t = (v − u) ÷ a = (15 − 25) ÷ (−1.0) = 10 s.
Common mistakeUsing s = ut + ½at² first is a dead end because both a and t are unknown; v² = u² + 2as needs no time.
44Short answer
Show why a speed of 1 m/s is the same as 3.6 km/h.
Show answer
Model answer: 1 m/s means 1 m every second, so in one hour (3600 s) the body travels 3600 m = 3.6 km. So 1 m/s = 3.6 km/h.
Common mistakeLearners often remember '3.6' but not which way to use it; working it out from 3600 s in an hour shows it.
45Short answer · ★ Challenge
Speed governors limit some buses in Rwanda to 60 km/h. With a braking deceleration of 6 m/s², how much shorter is the braking distance from 60 km/h than from 80 km/h?
Show answer
Model answer: 60 km/h = 16.7 m/s: s = v² ÷ 2a = 16.7² ÷ 12 = 23.1 m. 80 km/h = 22.2 m/s: s = 22.2² ÷ 12 = 41.2 m. Saving = 41.2 − 23.1 ≈ 18 m.
Common mistakeUsing km/h directly in v² = u² + 2as gives nonsense; convert to m/s first (÷ 3.6).
46Multiple choice
Without air resistance, how does the speed of a ball when it returns to the thrower's hand compare with the speed at which it was thrown upwards?
1. AIt returns more slowly
2. BThe two speeds are equal
3. CIt returns faster
4. DIt returns with zero speed
Show answer
Answer: B. The two speeds are equal
The motion down is the mirror image of the motion up: it gains back exactly the speed it lost.
Common mistakeChoosing 'more slowly' assumes the ball loses speed for good; without air resistance no energy is lost.
47Short answer · ★ Challenge
Two 5-space sections of a tape from a 50 Hz ticker-timer are 2.0 cm and 6.0 cm long. The middle of the second section was made 0.4 s after the middle of the first. Find the acceleration of the trolley.
Show answer
Model answer: Each section takes 5 × 0.02 = 0.1 s. v₁ = 2.0 ÷ 0.1 = 20 cm/s; v₂ = 6.0 ÷ 0.1 = 60 cm/s. a = (60 − 20) ÷ 0.4 = 100 cm/s² = 1.0 m/s².
Common mistakeDividing the change in velocity by 0.1 s (the time of one section) instead of 0.4 s (the time between the two sections) gives an answer four times too big.
48Short answer · ★ Challenge
A car starts from rest, accelerates at 2 m/s² for 5 s, moves at steady speed for 10 s, then decelerates at 5 m/s² for 2 s and stops. Describe its acceleration–time graph and check that the car really stops.
Show answer
Model answer: From 0 to 5 s: a horizontal line at +2 m/s². From 5 to 15 s: a line along the time axis (a = 0). From 15 to 17 s: a horizontal line at −5 m/s², below the axis. Check: Δv = 2 × 5 − 5 × 2 = 10 − 10 = 0, so the car ends at rest.
Common mistakeDrawing the steady-speed part at a height of 10 confuses velocity with acceleration; at steady speed a = 0.
49Multiple choice
On a displacement–time graph, which line shows a car that starts from rest and speeds up uniformly?
1. AA straight line through the origin
2. BA curve that gets less steep as time goes on
3. CA curve that gets steeper as time goes on
4. DA horizontal line above the time axis
Show answer
Answer: C. A curve that gets steeper as time goes on
s = ½at² grows faster and faster, so the gradient (velocity) keeps increasing: the curve gets steeper.
Common mistakeChoosing the straight line through the origin is tempting, but that shows constant velocity, not acceleration.
50Short answer · ★ Challenge
The velocity–time graph of a lift going up a Kigali office block: it rises uniformly from 0 to 3 m/s in 2 s, stays at 3 m/s until t = 10 s, then falls uniformly to 0 at t = 13 s. Find the total height risen and the deceleration at the end.
Show answer
Model answer: Height = area = ½ × 2 × 3 + 8 × 3 + ½ × 3 × 3 = 3 + 24 + 4.5 = 31.5 m. Deceleration = 3 ÷ 3 = 1 m/s².
Common mistakeUsing 10 s instead of 8 s for the steady part (it runs from t = 2 s to t = 10 s) gives too large an area.