Questions on motion graphs and the SUVAT equations

Question 1

Distinguish between the average velocity and the instantaneous velocity of a moving object.

Accepted Answer

Average velocity is total displacement divided by total time (1). Instantaneous velocity is the rate of change of displacement at one instant, that is the gradient of the displacement–time graph at that point (1).

Question 2

State what the gradient and the area under a velocity–time graph each represent.

Accepted Answer

The gradient represents acceleration (1); the area between the line and the time axis represents displacement (1).

Question 3

On a displacement–time graph, state what is represented by (a) the gradient at a point, and (b) a horizontal section of the line.

Accepted Answer

(a) The instantaneous velocity at that point (1). (b) The object is stationary; its displacement is not changing (1).

Question 4

An object accelerates uniformly from rest to 15 m s⁻¹ in 5.0 s, then travels at a constant 15 m s⁻¹ for a further 10 s. Using the area under the velocity–time graph, calculate the total distance travelled.

Accepted Answer

Distance = ½ × 5.0 × 15 (area of the triangle) (1) + 15 × 10 (area of the rectangle) (1) Distance = 188 m (1)

Question 5

The velocity–time graph of a journey is a straight line sloping upwards from a non-zero starting velocity. Explain how you would find the acceleration and the displacement from this graph.

Accepted Answer

The acceleration is the gradient of the line, change in velocity divided by change in time (1). The displacement is the area of the trapezium between the line and the time axis (1), which equals ½(u + v)t (1).

Question 6

A ball is released from rest, falls to a hard floor and bounces repeatedly, each bounce reaching a smaller height than the last, until it comes to rest. Air resistance is negligible while the ball is in flight. Taking downward velocity as positive, describe fully the velocity–time graph of the motion from release until the ball stops bouncing.

Accepted Answer

During every flight the graph is a straight line of gradient +g (9.81 m s⁻²), because gravity is the only force acting (1). Every up and every down section has this same gradient (1). From release the velocity rises from zero to its largest positive value just before the first impact (1). At each bounce the line is almost vertical: the velocity reverses from positive (downwards) to negative (upwards) in a very short contact time (1). Each rebound speed is smaller than the impact speed before it, because energy is lost in the bounce, so the peaks of the graph shrink (1). The velocity passes through zero at the top of each bounce, and the graph ends at v = 0 when the ball comes to rest (1).

Question 7

A car moves along a straight road for 20 s. Motion in one direction is taken as positive. Determine the displacement of the car from its starting point at the end of the 20 s, and the total distance it has travelled.

Accepted Answer

Displacement is the area between the line and the time axis, and area below the axis counts as negative because the car has reversed (1)
Above the axis the area is ½ × 6.0 × 12 + 6.0 × 12 + ½ × 6.0 × 12 = 36 + 72 + 36 = 144 m (1)
The line crosses the axis at t = 18 s, so the area below it is ½ × 2.0 × 4.0 = 4.0 m, and the displacement is 144 − 4.0 = 140 m (1)
Distance takes no account of direction, so the last 4.0 m is added rather than subtracted, giving 148 m (1)

Question 8

A car starts from rest and accelerates uniformly at 2.0 m s⁻² for 6.0 s. Calculate its final velocity.

Accepted Answer

v = u + at = 0 + 2.0 × 6.0 (1) v = 12 m s⁻¹ (1)

Question 9

A tram slows uniformly from 12.0 m s⁻¹ to 3.0 m s⁻¹ in 6.0 s. Calculate its acceleration.

Accepted Answer

a = Δv/Δt = (3.0 − 12.0)/6.0 (1) = −1.5 m s⁻², a deceleration of 1.5 m s⁻² (1).

Question 10

State the condition that must be satisfied for the SUVAT equations to be valid.

Accepted Answer

The acceleration must be constant (uniform) throughout the motion (1).

Question 11

A car travelling at 8.0 m s⁻¹ accelerates uniformly at 1.5 m s⁻² for 4.0 s. Calculate (a) its final velocity and (b) the distance travelled in this time.

Accepted Answer

(a) v = u + at = 8.0 + 1.5 × 4.0 (1) v = 14 m s⁻¹ (1) (b) s = ut + ½at² = 8.0 × 4.0 + ½ × 1.5 × 4.0² (1) s = 44 m (1)

Question 12

A car travelling at 20 m s⁻¹ brakes uniformly and stops in a distance of 40 m. Calculate (a) its deceleration and (b) the time taken to stop.

Accepted Answer

(a) v² = u² + 2as gives a = −(20²)/(2 × 40) (1) a = −5.0 m s⁻², a deceleration of 5.0 m s⁻² (1) (b) t = v/|a| = 20/5 (1) t = 4.0 s (1)

Question 13

A stone is dropped from rest and falls freely. Taking g = 9.81 m s⁻², calculate its velocity and the distance fallen after 2.0 s.

Accepted Answer

v = gt = 9.81 × 2.0 = 19.6 m s⁻¹ (1) s = ½gt² = ½ × 9.81 × 2.0² (1) s = 19.6 m (1)

Question 14

A student measures a friend's reaction time by dropping a ruler between the friend's open fingers. The friend catches the ruler after it has fallen 28 cm from rest. Taking g = 9.81 m s⁻², show that the reaction time is about 0.24 s.

Accepted Answer

From rest, s = ½gt², so t = √(2s/g) (1). t = √(2 × 0.28/9.81) (28 cm converted to 0.28 m) (1) = 0.239 s ≈ 0.24 s (1).

Question 15

A skateboarder is moving at 1.2 m s⁻¹ at the top of a ramp. She accelerates uniformly down the ramp at 2.5 m s⁻² over a distance of 4.2 m. Calculate her speed at the bottom of the ramp.

Accepted Answer

v² = u² + 2as (1) = 1.2² + 2 × 2.5 × 4.2 = 22.4 (1), so v = 4.7 m s⁻¹ (1).

Question 16

A goods train travels at 90 km h⁻¹. The driver applies the brakes and the train decelerates uniformly at 0.75 m s⁻². Calculate the minimum distance the train travels before stopping.

Accepted Answer

u = 90 km h⁻¹ = 90 000/3600 = 25 m s⁻¹ (1). With v = 0: v² = u² + 2as (1), so s = 25²/(2 × 0.75) (1) = 417 m ≈ 420 m (1).

Question 17

A train starts from rest and accelerates uniformly at 0.50 m s⁻² for 20 s. The driver then brakes and the train decelerates uniformly to rest in a further 100 m. Calculate (a) the maximum velocity of the train, (b) the deceleration while braking, and (c) the total distance travelled.

Accepted Answer

(a) v = u + at = 0 + 0.50 × 20 = 10 m s⁻¹ (1)
(b) v² = u² + 2as gives a = −10²/(2 × 100) (1)
a = −0.50 m s⁻², a deceleration of 0.50 m s⁻² (1)
(c) Accelerating: s = ½at² = ½ × 0.50 × 20² = 100 m (1)
Total = 100 + 100 = 200 m (1)

Question 18

A ball is thrown vertically upwards at 15 m s⁻¹. Taking g = 9.81 m s⁻² and ignoring air resistance, calculate (a) the maximum height reached and (b) the time taken to reach it.

Accepted Answer

(a) At the top v = 0, so h = u²/2g = 15²/(2 × 9.81) (1) h = 11.5 m (1) (b) t = u/g = 15/9.81 (1) t = 1.53 s (1)

Question 19

A delivery van passes a stationary motorcycle courier, travelling at a constant 16 m s⁻¹ along a straight road. At the instant it passes, the courier sets off after it with a uniform acceleration of 2.0 m s⁻². (a) Show that the courier draws level with the van 16 s after setting off. (b) Go on to calculate the distance travelled and the courier's speed at that moment.

Accepted Answer

(a) Equal displacements when the courier draws level: 16t = ½ × 2.0 × t² (1), so t = 16 s (1). (b) s = 16 × 16 = 256 m (1). v = u + at = 2.0 × 16 (1) = 32 m s⁻¹ (1).

Question 20

A car travels at 20 m s⁻¹ towards a pelican crossing. When the car is 42 m from the stop line, the lights change and the driver decides to stop. The driver's reaction time is 0.60 s and the car's maximum deceleration is 5.0 m s⁻². Deduce whether the car can stop before the stop line.

Accepted Answer

Thinking distance = 20 × 0.60 = 12 m (1). Braking distance = v²/2a = 20²/(2 × 5.0) = 40 m (1). Stopping distance = 12 + 40 = 52 m (1). 52 m is greater than 42 m, so the car cannot stop before the line (1).

Question 21

Derive the four constant-acceleration equations from the definition of acceleration and the definition of average velocity. State which step of your derivation uses the condition that the acceleration is constant.

Accepted Answer

With a constant acceleration the average and the instantaneous rate of change of velocity are the same, so a = (v − u)/t, which rearranges to v = u + at (1). Displacement is average velocity × time, and under a constant acceleration the average velocity is the mean of the end values, ½(u + v), giving s = ½(u + v)t (1). Substituting the first result into the second: s = ½(u + u + at)t = ½(2u + at)t (1), which multiplies out to s = ut + ½at² (1). Eliminating t instead: t = (v − u)/a, so s = ½(u + v)(v − u)/a, and multiplying both sides by 2a gives 2as = (v + u)(v − u) = v² − u², that is v² = u² + 2as (1). The condition is used in the second step, where the average velocity is taken as the mean of the first and last values; that is false as soon as the acceleration varies (1).

Questions on motion graphs and the SUVAT equations

Reading the motion graphs

The constant-acceleration equations