Velocity-Time Graphs Questions | Worksheets and Revision | MME

Velocity-Time Graphs Questions, Worksheets and Revision

Level 6 Level 7

Velocity-Time Graphs

A velocity-time graph(or speed-time graph) is a way of visually expressing a journey

We are going to be using velocity-time graphs to find two things, primarily: total distance, and acceleration.

Make sure you are happy with the following topics before continuing:

Velocity-Time Graphs – Key things to remember:

With speed on the $y$-axis and time on the $x$-axis, a speed-time graph tells us how someone/something’s speed has changed over a period of time.

1) The gradient of the line = Acceleration
3) Flat section means steady velocity (NOT STOPPED)
4) Area under the graph = Distance travelled

Example 1 – Acceleration and Distance travelled

The speed-time graph below describes a $50$-second car journey.

Work out the total distance travelled by the car and work out the maximum acceleration of the car during the $50$ seconds.

[4 marks]

Lets analyse the graph:

A) The car accelerated from $0$ to $15$m/s over the first $10$ seconds (because the line is straight, the acceleration is constant).

B) The line is flat, meaning the car’s speed did not change for $10$ seconds – meaning it was moving at a constant speed.

C) The car accelerated up to $25$m/s over the next $10$ seconds,

D) Finally it spent the last $20$ seconds decelerating back down to $0$m/s.

Work out the total distance travelled by the car:

Area under the graph = Distance travelled

To work out the area under this graph, we will break it into 4 shapes: $A$, $B$, $C$, and $D$ -two triangles, a rectangle, and a trapezium are all shapes that we can work out the area of. So, we get

$\text{A}=\dfrac{1}{2}\times10 \times15= 75\text{m}$

$\text{B}=10\times15=150\text{m}$

$C=\dfrac{1}{2}(15+25)\times10=200 \text{m}$

$D=\dfrac{1}{2}\times20\times25=250\text{m}$

Total distance travelled:

$75+150+200+250=675\text{m}$

Work out the maximum acceleration:

The gradient of the line = Acceleration

First lest find the gradient of the each section.

Section $\bf{A}$

$\text{acceleration between 0s and 10s = gradient}=\dfrac{15-0}{10-0}=1.5\text{m/s}^2$

Section $\bf{B}$

This section is flat, meaning meaning the acceleration will be $0$

Section $\bf{C}$

$\text{acceleration between 20s and 30s = gradient}=\dfrac{25-15}{30-20}=1\text{m/s}^2$

Section $\bf{D}$

$\text{acceleration between 30s and 50s = gradient}=\dfrac{0-25}{50-30}=\dfrac{-25}{20} = -1.25\text{m/s}^2$

Section $\bf{A}$ has the largest acceleration so is the maximum acceleration is $1.5\text{m/s}^2$.

Example 2

Below is a speed-time graph of the first $4$ seconds of someone running a race.

Work out the average acceleration over the $4$ seconds and work out the instantaneous acceleration $2$ seconds in.

we know:

The gradient of the line = Acceleration

Work out the average acceleration over the $4$ seconds

To work out the average acceleration over the 4 seconds, we will draw a line from where the graph is at $0$s to where the graph is at $4$s and find the gradient of it.

So, we get that average acceleration to be,

$\text{gradient}=\dfrac{6-0}{4-0}=1.5\text{ m/s}^2$

work out the instantaneous acceleration $2$ seconds in

To so this we will draw a tangent to the line after $2$ seconds and work out the gradient of that. This is shown above.

Then, we get the instantaneous acceleration to be,

$\text{gradient}=\dfrac{5.8-3.2}{3.5-1.0} = 1.04\text{ m/s}^2\text{ (3 s.f.)}$

Example Questions

So, we will firstly draw a straight from the origin to (12, 4), since after 12 seconds, it’s reached 4 m/s. Then, for the next part we’re told the deceleration is 0.1$\text{m/s}^2$ for 20 seconds. So, if the speed decreases by 0.1 every second, after 20 seconds it will be

$0.1 \times 20=2\text{m/s}$

Therefore, by 32 seconds in the speed is 2 m/s, so we will draw a straight line from (12, 4) to (32, 2). Finally, a constant speed will be represented by a flat line that goes until the 50 second point, still at 2 m/s. The result should look like the graph below.

We need to find the area underneath the graph. To do this, we will split it up into shapes we know how to calculate the area of, as seen below.

A is a triangle, B and C are trapeziums, and D is a rectangle. So, we get

$\text{A}=\dfrac{1}{2}\times10\times15=75\text{m}$

$\text{B}=\dfrac{1}{2}\times(10+15)\times5=62.5\text{m}$

$\text{C}=\dfrac{1}{2}\times(10+20)\times5=75\text{m}$

$\text{D}=30\times20 = 600\text{ m}$

Therefore, the total distance travelled by the cyclist is

$75+62.5+75+600=812.5\text{m}$

In order to determine the average acceleration we draw a line from the origin to the endpoint of the graph. The average acceleration sis given by the gradient of this line.

Hence the average acceleration is,

$\text{gradient}=\dfrac{4-0}{50-0}= 0.08\text{ m/s}^2$

Level 6-9

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