# Velocity-Time Graphs

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## 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.

There are 5 key skills you need to learn

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

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## 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
2) Negative gradient = Deceleration
3) Flat section means constant velocity (NOT STOPPED)
4) Area under the graph = Distance travelled

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## Skill 1: Describing a graph

One Skill you will need learn is describing a velocity time graph.

Example: The speed-time graph shows a $50$-second car journey. Describe the $50$ second journey.

Step 1: Split the graph up into distinct sections, these can be seen in the image as $A, B, C$ and $D$.

Step 2: In detail describe each part of the journey, ensuring to use numerical values throughout.

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

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

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

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

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## Skill 2: Calculating Acceleration

Acceleration is calculated as the change in speed over time.

Example: The speed-time graph shows a $50$-second car journey, find which section of the graph has the greatest acceleration.

We know,

The gradient of the line = Acceleration

We must find the gradient of the each section.

Section $\bf{A}$:  Acceleration between $0$s and $10$s $=$ gradient$=\dfrac{15-0}{10-0}=1.5$ m/s$^2$

Section $\bf{B}$: This section is flat, meaning the acceleration will be $0$

Section $\bf{C}$: Acceleration between $20$s and $30$s $=$ gradient $=\dfrac{25-15}{30-20}=1$ m/s$^2$

Section $\bf{D}$: Acceleration between $30$s and $50$s $=$ gradient $=\dfrac{0-25}{50-30}=\dfrac{-25}{20} = -1.25$ m/s$^2$

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

Note: units of acceleration are expressed in distance/time$\bold{^2}$, which in this case is m/s$\bold{^2}$.

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## Skill 3: Calculating total distance travelled

Calculating the total distance travelled is one of the most common exam questions you may see.

Example: The speed-time graph shows a $50$-second car journey, Calculate the total distance travelled over the $50$ seconds.

we know,

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 $D$.

This gives two triangles, a rectangle, and a trapezium, which are all shapes that we can work out the area of.

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

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

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

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

Total distance travelled:

$75+150+200+250=675$ m

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## Skill 4: Average of curved graphs

Finding the average gradient, is the gradient over a length of time.

Example: A speed-time graph of the first $4$ seconds of someone running a race is shown.

Calculate the average acceleration over the $4$ seconds.

We know:

The gradient of the line = Acceleration

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 the average acceleration to be,

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

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## Skill 5: Instantaneous gradient of a curve

Finding the instantaneous gradient, is the gradient of the tangent at a point.

Example: A speed-time graph of the first $4$ seconds of someone running a race is shown.

Calculate the instantaneous acceleration $2$ seconds in.

To do 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,

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

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## Velocity-Time Graphs 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 $0.1$ m/s$^2$ for $20$ seconds. So, if the speed decreases by $0.1$ every second, after $20$ seconds it will be

$0.1 \times 20=2$ 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.

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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

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

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

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

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

Therefore, the total distance travelled by the cyclist is

$75+62.5+75+600=812.5$ m

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In order to determine the average acceleration, we draw a line from the origin to the endpoint of the graph, as seen below. The average acceleration is given by the gradient of this line.

Hence the average acceleration is,

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

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## Velocity-Time Graphs Worksheet and Example Questions

### (NEW) Velocity-Time Graphs Exam Style Questions - MME

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