Velocity/time graphs

Question 1

What does the slope/gradient of the graph tell you?

Answer 1

The acceleration of the vehicle.

Question 2

What is the equation for working out the acceleration?

Answer 2

Acceleration = change in velocity ÷ change in time

Question 3

Which axis is time plotted on?

Answer 3

The x-axis.

Question 4

What does the area under the graph tell you?

Answer 4

The distance travelled.

Question 5

How do you work out the areas underneath the graph?

Answer 5

Rectangle: base x height

Triangle: 1/2 x base x height

Question 6

A car travels at a constant speed of 8 m/s for 12 s before accelerating at 1.5 m/s² for the next 6 s.

a) Sketch a velocity/time graph for the car.
b) Calculate the total distance travelled by the car.

Answer 6

a) correct axes shown: velocity on y-axis and time on x-axis,
horizontal line at 8 m/s shown for 12 s,
positive linear gradient of 1.5 shown for the next 6 seconds,
final velocity of 17 m/s shown after 18 s.

b) distance travelled = area under graph
distance travelled in first 8 s = 8 m/s × 12 s = 96 m
distance travelled in second part of journey = 8 m/s × 6 s + ½ × 9 m/s × 6 s = 48 m + 27 m = 75 m
total distance = 96 m + 75 m = 171 m

Question 7

Draw this graph:

A train is stationary from 0 - 10 seconds.
It accelerates until 30 seconds, where it reaches a velocity of 20 m/s.
It travels at a constant speed for 20 more seconds.
It then decelerates for 20 seconds by -10 m/s in total.
It then accelerates for 10 seconds by 30 m/s in total.
It then travels at a constant speed for 10 seconds.

Use the velocity/time graph of the train to calculate:

a) the distance travelled after 50 s.

b) the acceleration of the train between 50 s and 70 s.
a) area beneath the graph = [½ × 20 × (30 − 10)] + [20 × (50 − 30)]

= 200 m + 400 m = 600 m.

b) Change in velocity = 10 m/s - 20 m/s = -10 m/s
Change in time = 70 s - 50 s = 20 s

Acceleration = -10 m/s ÷ 20 s = -0.5 m/s²

Answer 7

a) area beneath the graph = [½ × 20 × (30 − 10)] + [20 × (50 − 30)]

= 200 m + 400 m = 600 m.

b) Change in velocity = 10 m/s - 20 m/s = -10 m/s
Change in time = 70 s - 50 s = 20 s

Acceleration = -10 m/s ÷ 20 s = -0.5 m/s²

Question 8

State what can be determined from the (a) gradient and (b) area beneath a velocity/time graph.

Answer 8

a) The gradient of a velocity/time graph gives acceleration.

b) Area beneath a velocity/time graph gives distance travelled.