IGCSEScienceOxford AQACombined Science Double AwardPhysicsRevision NotesForces & Their EffectsResultant ForcesVelocity-Time Graphs

Velocity-Time Graphs (Oxford AQA IGCSE Combined Science Double Award)

Revision Note

Leander Oates
Oxford AQA Combined Science Double Award: PhysicsOxford AQA Physics

Calculating Acceleration from a Velocity-Time Graph

Extension Tier only

  • The gradient of the line on a velocity-time graph represents the magnitude of acceleration

    • A steep gradient means large acceleration (or deceleration)

      • The object's velocity changes very quickly

    • A gentle gradient means small acceleration (or deceleration)

      • The object's velocity changes very gradually

    • A horizontal line means the acceleration is zero

      • The object is moving with a constant velocity

      • A constant velocity means a constant speed in a straight line

      • A horizontal line at v = 0 shows a stationary object

Interpreting gradients on a Velocity-Time graphs

Different gradients on a velocity-time graph for IGCSE & GCSE Physics revision notes
This image shows how to interpret the gradients of a velocity-time graph

  • The acceleration of an object can be calculated from the gradient of a velocity-time graph

 acceleration space equals space gradient space equals space fraction numerator increment y over denominator increment x end fraction

  • Here, Δy is change in velocity and Δx is change in time

Calculating the gradient of a Velocity-Time graph

Annotated velocity-time graph for IGCSE & GCSE Physics revision notes
To find the gradient, you calculate ∆y, the change in velocity, over ∆x, the change in time

Worked Example

A cyclist is training for a tournament.

The velocity-time graph below shows the cyclist's motion as they cycle along a flat, straight road.

A graph showing five line segments. Segment A is a horizontal line at v=0 between 0 and 5 seconds. Segment B is a straight ascending line as the object's velocity increases from 0 to 5 metres per second between the time interval of 5 to 10 seconds. Segment C is horizontal at v = 5 metres per second from t = 10 seconds to t = 13 seconds. Segment D ascends sharply again as the velocity increases from 5 to 10 metres per second over the time interval between 13 and 15 seconds. Segment E is a horizontal line at v = 10 metres per second from t = 15 seconds to t = 20 seconds

(a) In which section of the velocity-time graph is the cyclist's acceleration the largest?

(b) Calculate the cyclist's acceleration between 5 and 10 seconds.

Answer:

Part (a)

Step 1: Recall what the gradient of a velocity-time graph shows

  • The gradient of a velocity-time graph indicates the magnitude of acceleration

  • Therefore, the only sections of the graph where the cyclist is accelerating are section B and section D

  • Sections A, C, and E are flat – in other words, the cyclist is moving at a constant speed (i.e. not accelerating)

Step 2: Identify the section with the steepest gradient

  • Section D of the graph has the steepest gradient

  • Hence, the greatest acceleration is shown in section D

Part (b)

Step 1: Draw a large gradient triangle at the appropriate section of the graph

  • A gradient triangle is drawn for the time interval between 5 and 10 seconds below:

The graph is annotated with a gradient triangle showing the change in velocity is zero to 5 metres per second for the time interval t = 5 seconds to t = 10 seconds (segment B)

Step 3: Calculate the gradient

acceleration space equals space gradient space equals fraction numerator increment y over denominator increment x end fraction

a space equals fraction numerator space 5 over denominator 5 end fraction

a space equals space 1 space straight m divided by straight s squared

  • Therefore, the cyclist accelerated at 1 m/s2 between 5 and 10 seconds

Examiner Tip

Use the entire gradient line, where possible, to calculate the gradient. Examiners tend to award credit if they see a large gradient triangle used - so remember to draw the lines directly on the graph itself!

Calculating Distance from a Velocity-Time Graph

Extension Tier only

  • The distance travelled by an object can be found by determining the area beneath a velocity-time graph

Calculating the area under a Velocity-Time graph

The area under a velocity time graph can be calculated using the area of a triangle for a sloped line and the area of a rectangle for a horizontal line. For IGCSE & GCSE Physics revision notes
The distance travelled can be found from the area beneath the graph

  • If the area beneath the graph forms a triangle (the object is accelerating or decelerating) then the area can be determined using the formula:

 area space equals space 1 half cross times space base space cross times space height

  • If the area beneath the graph is a rectangle (constant velocity) then the area can be determined using the formula:

 area space equals space base space cross times space height

Worked Example

The velocity-time graph below shows a car journey which lasts for 160 seconds.

A velocity-time graph with 4 segments. The first segment is an upward sloping line from v = 0 to v = 17.5 metres per second over the time interval t = 0 to t = 40 seconds. The second is a horizontal line at v = 17.5 metres per second for the time interval t = 40 seconds to t = 70 seconds. The third is an upward sloping line from v = 17.5 metres per second to v = 25 metres per second for the time interval t = 70 seconds to t = 90 seconds. The final line segment is a downward sloping line from v = 25 metres per second to v = 0 over the time interval t = 90 seconds to t = 160 seconds.

Calculate the total distance travelled by the car on this journey.

Answer:

Step 1: Recall that the area under a velocity-time graph represents the distance travelled

  • To calculate the total distance travelled, the total area underneath the line must be determined

Step 2: Identify each enclosed area

  • In this example, there are five enclosed areas under the line

  • These can be labelled as areas 1, 2, 3, 4 and 5, as shown in the image below:

The same graph has been annotated to show area under each line segment in triangles and rectangles. Sloping lines are calculated using triangles and flat lines are calculated using rectangles. However, segment 3's area consists of a rectangle and a triangle on top of it.

Step 3: Calculate the area of each enclosed shape under the line

  • Area 1 = area of a triangle

Area space 1 space equals 1 half space cross times space base space cross times space height

Area space 1 space equals space 1 half space cross times space 40 space cross times space 17.5

Area space 1 space equals space 350 space straight m

  • Area 2 = area of a rectangle

Area space 2 space equals space base space cross times space height

Area space 2 space equals space 30 space cross times space 17.5

Area space 2 space equals space 525 space straight m

  • Area 3 = area of a triangle

Area space 3 space equals space 1 half space cross times space base space cross times space height

Area space 3 space equals space 1 half space cross times space 20 space cross times space 7.5

Area space 3 space equals space 75 space straight m

  • Area 4 = area of a rectangle

Area space 4 space equals space base space cross times space height

Area space 4 space equals space 20 space cross times space 17.5

Area space 4 space equals space 350 space straight m

  • Area 5 = area of a triangle

Area space 5 space equals space 1 half space cross times space base space cross times space height

Area space 5 space equals space 1 half space cross times space 70 space cross times 25

Area space 5 space equals space 875 space straight m

Step 4: Calculate the total distance travelled by finding the total area under the line

  • Add up each of the five areas enclosed:

total space distance space space equals space Area space 1 space plus space Area space 2 space plus space Area space 3 space plus space Area space 4 space plus space Area space 5

total space distance space equals space 350 space plus space 525 space plus space 75 space plus space 350 space plus space 875

total space distance space equals space 2175 space straight m

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    Leander Oates

    Author: Leander Oates

    Expertise: Physics

    Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.

    Caroline Carroll

    Author: Caroline Carroll

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    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.