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GCSE · Physics

IG.P2 Distance-Time and Velocity-Time Graphs

EDEXCEL 4PH1 · Calculator allowed · about 65 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Read off from the distance-time graph of a pedestrian walking along a straight path: at 0 minutes the distance is 0 m; at 5 minutes the distance is 300 m. State the average speed of the pedestrian between 0 and 5 minutes in m/s.

(Total for Question 1 is 3 marks)

A car travels with constant speed. On a distance-time graph its line is straight and passes through (2 s, 10 m) and (6 s, 34 m). Calculate the speed of the car in m/s.

(Total for Question 2 is 2 marks)

Describe the motion shown by a distance-time graph that is a horizontal straight line at 20 m from the origin for all times between 0 s and 10 s.

(Total for Question 3 is 2 marks)

A cyclist starts from rest and accelerates uniformly so the distance-time graph is a curve. Explain how the shape of the curve shows the cyclist is accelerating.

(Total for Question 4 is 3 marks)

Read off from a velocity-time graph a vehicle that travels at 12 m/s from t = 0 s to t = 4 s, then decelerates linearly to 0 m/s at t = 8 s. Determine the acceleration during the deceleration phase, in m/s².

(Total for Question 5 is 3 marks)

A velocity-time graph shows a car moving at 10 m/s for 6 s, then accelerating uniformly to 22 m/s over the next 4 s. Calculate the distance travelled during the acceleration interval (from t = 6 s to t = 10 s) using the area under the velocity-time graph. Give your answer in metres.

(Total for Question 6 is 4 marks)

Sketch on a blank axes a velocity-time graph for a toy car that starts from rest, accelerates uniformly to 5 m/s in 2 s, then continues at 5 m/s for 3 s, and finally brakes uniformly to rest in 1 s. Label the time points t = 0 s, 2 s, 5 s and 6 s and the velocities 0 and 5 m/s. No precise drawing required, but shapes and labels must be correct.

(Total for Question 7 is 3 marks)

Using the graph sketched in Question 7, calculate the total distance travelled by the toy car from t = 0 to t = 6 s by finding the area under the velocity-time graph. Give your answer in metres.

(Total for Question 8 is 3 marks)

A runner completes a two-stage journey shown on a velocity-time graph: first stage, constant speed 4 m/s for 10 s; second stage, accelerates uniformly to 12 m/s over 5 s. Calculate the total distance run during the whole 15 s interval by using areas under the velocity-time graph. Show steps and give units.

(Total for Question 9 is 4 marks)

A distance-time graph shows an object moving away from the origin with constant speed, covering 90 m in 30 s. Read off the speed from the graph in m/s.

(Total for Question 10 is 2 marks)

A lorry is shown on a velocity-time graph as follows: accelerates uniformly from 0 to 8 m/s in 4 s, continues at 8 m/s for 6 s, then decelerates uniformly to 2 m/s in 2 s. Calculate (a) the acceleration during the initial 4 s, and (b) the distance covered during the whole 12 s journey by using the areas under the graph. Give units.

(a)Calculate the acceleration during the initial 4 s (from 0 to 8 m/s).(2)

(b)Calculate the distance covered during the whole 12 s journey using areas under the velocity-time graph.(2)

(Total for Question 11 is 4 marks)

Describe the motion of a vehicle shown on a distance-time graph that is a straight line with increasing gradient for the first 5 s, then a straight horizontal line for the next 5 s, and finally a straight line returning to the origin over 5 s. Use words describing speed and direction.

(Total for Question 12 is 2 marks)

A velocity-time graph shows a toy train moving at 3 m/s for 4 s, then accelerating uniformly to 9 m/s over 3 s, then maintaining 9 m/s for 5 s. Calculate (a) the acceleration during the 3 s interval, and (b) the distance covered in the entire 12 s. Show working and include units.

(a)Calculate the acceleration during the 3 s interval (from 3 m/s to 9 m/s).(1)

(b)Calculate the distance covered in the whole 12 s.(2)

(Total for Question 13 is 3 marks)

On a velocity-time graph a cyclist travels at -2 m/s (negative sign indicates towards the origin) for 5 s, then reverses direction and accelerates to +6 m/s over 4 s. Calculate the change in velocity and the average acceleration during the 4 s interval. Give units.

(Total for Question 14 is 3 marks)

A vehicle is driven so that its velocity-time graph between t = 0 and t = 9 s is as follows: from 0 to 3 s constant at 0 m/s (stationary), from 3 to 6 s accelerates uniformly to 15 m/s, from 6 to 9 s decelerates uniformly to 6 m/s. Calculate the total displacement from t = 3 s to t = 9 s using areas under the velocity-time graph. Give your answer in metres.

(Total for Question 15 is 3 marks)

Mark scheme · IG.P2 Distance-Time and Velocity-Time Graphs

Question 1

M1 forms change in distance over change in time fraction, (300 - 0) / (5 min)
M1 converts 5 minutes to seconds or converts result to m/s, e.g. 5 min = 300 s
A1 gives final answer 1.0 m/s cao
Answer: 1.0 m/s

Question 2

M1 uses gradient = change in distance / change in time = (34 - 10)/(6 - 2)
A1 gives 6 m/s cao
Answer: 6 m/s

Question 3

B1 states that distance does not change with time
B1 concludes the object is stationary at 20 m from the origin
Answer: The object is stationary; it stays at 20 m from the origin for the whole 0 s to 10 s interval.

Question 4

B1 notes the curve gets steeper with time
B1 links steeper gradient to increasing speed
B1 concludes speed is increasing so the cyclist is accelerating
Answer: The curve becomes steeper over time, so the gradient, which equals speed, increases. Therefore the cyclist's speed increases and the cyclist is accelerating.

Question 5

M1 uses gradient = change in velocity / change in time = (0 - 12)/(8 - 4)
A1 gives acceleration = -3.0 m/s² cao
B1 states negative sign means deceleration or slowing down
Answer: -3.0 m/s² (deceleration of 3.0 m/s²)

Question 6

M1 recognises area under v-t graph for the interval is a trapezium and uses area = (sum of parallel sides)/2 x height
M1 substitutes velocities and time: area = (10 + 22)/2 x 4
A1 evaluates area = 64 m cao
B1 states unit m with final answer
Answer: 64 m

Question 7

B1 shows a straight line from (0,0) to (2,5) representing uniform acceleration
B1 shows a horizontal line from (2,5) to (5,5) representing constant speed
B1 shows a straight line from (5,5) to (6,0) representing uniform braking to rest, and labels times and velocities
Answer: Sketch showing: straight line 0 to (2,5), horizontal to (5,5), straight down to (6,0); axes labelled and points marked.

Question 8

M1 splits area into triangle for 0-2 s, rectangle for 2-5 s, triangle for 5-6 s, or equivalent
M1 calculates areas: triangle 0-2 s = 0.5 x 2 x 5 = 5 m, rectangle 2-5 s = 3 x 5 = 15 m, triangle 5-6 s = 0.5 x 1 x 5 = 2.5 m
A1 gives total = 22.5 m cao with unit
Answer: 22.5 m

Question 9

M1 calculates distance in stage 1 as area of rectangle = 4 x 10
M1 recognises stage 2 area is trapezium or triangle+rectangle and uses area = (4 + 12)/2 x 5
A1 evaluates stage 1 = 40 m and stage 2 = 40 m cao
B1 gives total = 80 m with unit
Answer: 80 m

Question 10

M1 uses speed = distance / time = 90 / 30
A1 gives speed = 3.0 m/s cao
Answer: 3.0 m/s

Question 11

(a) M1 uses gradient = change in velocity / change in time = (8 - 0)/4
(a) A1 gives acceleration = 2.0 m/s² cao
(a) Answer: 2.0 m/s²
(b) M1 splits into triangle 0-4 s, rectangle 4-10 s, trapezium 10-12 s or equivalent and calculates areas
(b) A1 evaluates total distance = 16 + 48 + 10 = 74 m cao
(b) Answer: 74 m

Question 12

B1 states first period is speeding up (increasing speed) away from origin
B1 states second period is moving at constant speed, then final period is returning to origin meaning moving back towards start and covering decreasing distance
Answer: From 0-5 s the vehicle is accelerating and increasing its speed away from the origin. From 5-10 s it moves at constant speed. From 10-15 s it moves back towards the origin, its distance from the origin decreasing until it reaches the origin.

Question 13

(a) A1 gives acceleration = (9 - 3)/3 = 2.0 m/s² cao
(a) Answer: 2.0 m/s²
(b) M1 calculates areas for sections: rectangle 0-4 s = 3 x 4, trapezium 4-7 s = (3 + 9)/2 x 3, rectangle 7-12 s = 9 x 5
(b) A1 adds values: 12 + 18 + 45 = 75 m cao
(b) Answer: 75 m

Question 14

M1 calculates change in velocity = final - initial = 6 - (-2) = 8 m/s
M1 uses acceleration = change in velocity / time = 8 / 4
A1 gives acceleration = 2.0 m/s² cao and labels change in velocity 8 m/s
Answer: Change in velocity = 8 m/s; average acceleration = 2.0 m/s²

Question 15

M1 splits into trapezium 3-6 s with velocities 0 to 15 and time 3, and trapezium 6-9 s with velocities 15 to 6 and time 3, or uses triangle+rectangle forms
M1 calculates areas: 3-6 s area = (0 + 15)/2 x 3 = 22.5 m; 6-9 s area = (15 + 6)/2 x 3 = 31.5 m
A1 gives total displacement = 54.0 m cao
Answer: 54.0 m

Distance-Time and Velocity-Time Graphs - Worksheets, Questions and Revision