Paper Together: sit a paper in a room with your friends.See how it works
This page sets out the method for velocity–time graphs in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Velocity–time graphs topic page.
When the graph is made of straight sections, do this.
When you see 'use 4 strips of equal width to estimate the distance', do this.
When you see 'estimate the acceleration at ' on a curved graph, do this.
Time goes across. Velocity goes up. A point on the graph is a speed at a moment.
Acceleration is change in velocity divided by time. A downward line is a negative acceleration.
From rest up to a speed, the area is a triangle: half times time times final speed.
If the speed starts at u and ends at v, the area is the average of u and v, times time.
Split a graph into a rectangle and a triangle or trapezium, then add the areas.
If time is in seconds and velocity in m/s, area is metres.
A line from (0, 0) to (4, 8) rises 8 m/s in 4 seconds. The gradient is 8/4=2, so the acceleration is 2 m/s^2. Read the change from the graph, not the height alone. A line that starts at 8 m/s and is still at 8 m/s four seconds later has gradient 0. The object is moving quickly, but it is not accelerating.
A line that falls has a negative gradient. From 10 m/s down to 4 m/s in 3 seconds, the change in velocity is 4-10=-6, so the acceleration is (-6)/3=-2 m/s^2. If the question asks for the deceleration, the answer is 2 m/s^2. Say which quantity you have given. Mixing the sign with the word "deceleration" double-counts the slowing.
Method: 1. Read the change in velocity from the vertical axis. 2. Divide by the change in time from the horizontal axis. 3. A negative gradient is a negative acceleration. 4. If the question asks for deceleration, give that rate as a positive number.
Constant velocity is a rectangle. At 12 m/s for 5 seconds the area is 12× 5=60, so the distance is 60 metres. The acceleration on that section is 0. Units multiply as (m/s)×s=m. Write the unit with the number. A bare 60 can be marked as an unfinished answer when the question asked for a distance.
A straight acceleration from rest is a triangle. From rest to 8 m/s in 4 seconds the area is 1/2× 4× 8=16 metres. Forgetting the half counts the rectangle around the triangle and doubles the distance. A later steady section is a rectangle sitting on top of the time axis, not a second triangle. Add the pieces. Do not multiply the final velocity by the total time unless the velocity really was constant for the whole journey.
Method: 1. Split the region under the graph into triangles, rectangles and trapezia. 2. A triangle is half the base times the height. 3. A rectangle is base times height. Do not halve it. 4. Add the areas. The unit is metres when velocity is in m/s and time is in seconds.
Suppose the velocity rises from 0 to 10 m/s in 5 seconds, then falls from 10 m/s to 4 m/s in the next 3 seconds, then stays at 4 m/s for 7 seconds. The accelerations are 10/5=2, then (4-10)/3=-2, then 0. The total time is 5+3+7=15 seconds. Mark the breaks on the time axis before you find any area, so each shape has the right width.
The areas are a triangle 1/2× 5× 10=25, a trapezium 1/2(10+4)× 3=21, and a rectangle 4× 7=28. The distance is 25+21+28=74 metres. The middle shape is a trapezium because the two vertical sides are the parallel sides of lengths 10 and 4. Treating it as a triangle of height 10 or as a rectangle of height 4 both miss the sloping top.
Method: 1. Mark the time breaks before you calculate. 2. Find the acceleration of each straight section separately. 3. Match each section with its own area formula. 4. A sloping top between two heights is a trapezium, not a triangle of the taller height.
When the graph is curved, read the velocity at equal time steps and join the readings with straight lines. For widths of 2 seconds and velocities 0, 3, 5, 6, 6 the four trapezia have areas 2/2(0+3)=3, 2/2(3+5)=8, 2/2(5+6)=11 and 2/2(6+6)=12. The estimate is 3+8+11+12=34 metres.
The same total comes from the trapezium rule in one line: 2/2big(0+6+2(3+5+6)big)=34. The first and last heights appear once, and every height in between appears twice. This is an estimate, because the straight tops are not the curve. Say "estimate" if the graph was curved. An exact area is only available when the graph is made of straight sections and you have used those sections.
Method: 1. Read the velocity at equal time steps. 2. Each strip is a trapezium of width equal to the step. 3. Average the two heights and multiply by the width. 4. Say the total is an estimate when the original graph was a curve.
Real exam questions call these speed-time graphs, with speed on the vertical axis and time on the horizontal axis. Before you calculate anything, read the scale on each axis. A grid square is not always one unit, so count what each small square is worth. Then list the corner points of the graph as coordinates such as (0, 0), (40, 20) and (100, 20).
Units must match. If speed is in m/s then time must be in seconds, and the area gives metres. If the speed is in km/h and the time is in minutes, convert one of them first. To change km/h to m/s divide by 3.6, so 72 km/h is 20 m/s. Acceleration has units m/s^2, which you can write as metres per second squared.
Method: 1. Read the scale on both axes and write down the corner points. 2. Check the units and convert so speed and time agree. 3. Use the gradient for acceleration and the area for distance. 4. Write the unit with every answer: m/s^2 for acceleration and m for distance.
Tip: Divide by 3.6 to go from km/h to m/s, and multiply by 3.6 to go back.
A favourite exam question gives the total distance and asks for an unknown speed. The graph might rise from 0 to 6 m/s in 20 seconds, stay level for 60 seconds, and then fall to an unknown speed v over the last 20 seconds. The total distance is 490 m. Split the graph into shapes and write the area of each, using v in the last one.
The last section is a trapezium with parallel sides 6 and v and width 20, so its area is tfrac12(6+v)× 20=10(6+v). The other areas are 60 and 360. Setting the sum equal to 490 gives 60+360+10(6+v)=490, so 10(6+v)=70 and v=1. Form the equation, solve it, and check the answer by recalculating the total.
Method: 1. Split the graph into triangles, rectangles and trapezia. 2. Write the area of each shape, using a letter for the unknown value. 3. Add the areas and set them equal to the total distance. 4. Solve for the unknown and check by finding the total again.
Average speed over a whole journey is the total distance divided by the total time. It is not the mean of the highest and lowest speeds unless the acceleration is constant throughout. For a journey of 370 m taking 25 seconds the average speed is 370÷ 25=14.8 m/s. Find the total distance from the areas first, then divide.
To find when a certain speed is reached on a straight section, use the gradient. If the speed falls from 28 m/s to 10 m/s in 9 seconds, the deceleration is 18÷ 9=2 m/s^2. The speed is 20 m/s when it has dropped by 8, which takes 8÷ 2=4 seconds. Count the time from the start of that section, not from the start of the journey.
Method: 1. Find the total distance by adding the areas of all the sections. 2. Divide by the total time to get the average speed. 3. For a time on a sloping section, divide the change in speed by the gradient. 4. Add on the time at the start of that section if the question measures from the start of the journey.
On a curved graph the acceleration changes all the time. To estimate the distance, split the area into strips of equal width and treat each as a trapezium. With readings 0, 8, 12, 14, 15 every 5 seconds the estimate is tfrac12× 5×[(0+8)+(8+12)+(12+14)+(14+15)]=207.5 m. More strips give a better estimate. The more strips you use, the closer the estimate gets to the true area, but the exam usually fixes the number of strips for you.
To estimate the acceleration at one moment, draw a tangent to the curve at that time and work out its gradient. A question may ask whether the estimate is too big or too small. If the curve bends downwards, the straight tops of the trapezia lie below it, so the estimate is an under-estimate. If the curve bends upwards, the trapezia lie above it and the estimate is an over-estimate.
Method: 1. Draw vertical strips of equal width and read the speed at each edge. 2. Use the area of a trapezium for each strip, or sum the first and last readings with twice the others, times half the width. 3. For a gradient at a point, draw a tangent and use change in speed over change in time. 4. To say whether it is an under-estimate or over-estimate, look at which side of the curve the strip tops are on.
AQA-style and Edexcel-style questions. Yumio is not affiliated with, endorsed by or connected to AQA, Pearson Edexcel or any other exam board.