Motion Along a Straight Line (AQA A Level Physics): Revision Note

Exam code: 7408

Katie M

Written by: Katie M

Reviewed by: Tim

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Displacement, speed, velocity & acceleration

Scalar quantities

  • Scalar quantities only have a magnitude (size)

    • Distance: the total length between two points

    • Speed: the total distance travelled per unit of time

Vector quantities

  • Vector quantities have both magnitude and direction

    • Displacement: the distance of an object from a fixed point in a specified direction

    • Velocity: the rate of change of displacement of an object

    • Acceleration: the rate of change of velocity of an object

Equations for velocity and acceleration

  • The equation for velocity is

v = ∆s∆t

  • Where:

    • v = velocity (m s-1)

    • ∆s = change in displacement (m)

    • ∆t = change in time (s)

  • The equation for acceleration is

a = ∆v∆t

  • Where:

    • a = acceleration (m s-2)

    • ∆v = change in velocity (m s-1)

    • ∆t = change in time (s)

Worked Example

A car accelerates uniformly from rest to a speed of 150 km h–1 in 6.2 s. Calculate the magnitude of the acceleration of the car in m s–2.

[2]

Answer:

Step 1: Convert the speed from km h-1 to m s-1

150 km h−1 = 150 × 103 m h−1

3600 s = 1 h

150 × 1033600 = 41.67 m s−1 [1 mark]

Step 2: Calculate the acceleration

a = ∆v∆t

a = 41.676.2 = 6.7 m s−2 [1 mark]

Average & instantaneous speed

Instantaneous speed / velocity

  • The instantaneous speed (or velocity) is the speed (or velocity) of an object at any given point in time

  • This could be for an object moving at a constant velocity or accelerating

    • An object accelerating is shown by a curved line on a displacement – time graph

    • An accelerating object will have a changing velocity

  • To find the instantaneous velocity on a displacement-time graph:

    • Draw a tangent at the required time

    • Calculate the gradient of that tangent

Curved displacement-time graph with a red tangent drawn at t = 9 s. The tangent uses Δs = (10 − 0) m and Δt = (14 − 5) s, with axes s/m and t/s.
The instantaneous velocity is found by drawing a tangent on the displacement time graph

Average speed / velocity

  • The average speed (or velocity) is the total distance (or displacement) divided by the total time

  • To find the average velocity on a displacement-time graph, divide the total displacement (on the y-axis) by the total time (on the x-axis)

    • This method can be used for both a curved or a straight-line on a displacement-time graph

Worked Example

A cyclist travels a distance of 20 m at a constant speed then decelerates to a traffic light 5 m ahead. The whole journey takes 3.5 s. Calculate the average speed of the cyclist.

[2]

Answer:

Step 1: Write the average speed equation

Average speed = total distancetotal time

Step 2: Calculate the total distance

Total distance = 20 + 5 = 25 m [1 mark]

Step 3: Calculate the average speed

Average speed = 253.5 = 7.1 m s−1 [1 mark]

Uniform & non-uniform acceleration graphs

  • Three types of graph that can represent motion are displacement-time graphs, velocity-time graphs and acceleration-time graphs

Displacement-time graph

  • On a displacement-time graph:

    • The gradient (or slope) equals velocity

    • The y-intercept equals the initial displacement

    • A diagonal straight line represents a constant velocity

      • A positive slope represents motion in the positive direction

      • A negative slope represents motion in the negative direction

    • A curved line represents an acceleration

    • A horizontal line (zero slope) represents a state of rest

    • The area under the curve is meaningless

  • Remember the displacement-time graph can have positive or negative values on the displacement axis

    • However, a distance-time graph only has positive values

Three displacement–time graphs with displacement in metres against time in seconds. A straight line shows constant velocity, a curve steepening shows velocity increasing at a constant rate, and a sharply steepening curve shows increasing acceleration.
The gradient of a displacement-time graph gives the velocity

Velocity-time graph

  • On a velocity-time graph:

    • Slope equals acceleration

    • The y-intercept equals the initial velocity

    • A straight line represents uniform acceleration

      • A positive slope represents an increase in velocity (acceleration) in the positive direction

      • A negative slope represents an increase in velocity (acceleration) in the negative direction

    • A curved line represents the non-uniform acceleration

    • A horizontal line (zero slope) represents motion with constant velocity

    • The area under the curve equals the displacement or distance travelled

  • Remember the velocity-time graph can have positive or negative values on the displacement axis

    • However, a speed-time graph only has positive values

Four velocity–time graphs with v in metres per second and t in seconds. A horizontal line shows constant velocity, a straight rising line for increasing velocity, an upward-curving line for increasing acceleration, and a flattening curve for decreasing acceleration (deceleration).
The gradient of a velocity-time graph gives the acceleration

Acceleration-time graph

  • On an acceleration-time graph:

    • The slope is meaningless

    • The y-intercept equals the initial acceleration

    • A horizontal line (zero slope) represents an object undergoing constant acceleration

    • The area under the curve equals the change in velocity

Three labelled acceleration–time graphs with acceleration in metres per second squared and time in seconds. Zero acceleration shows constant velocity, a straight horizontal line shows constant positive acceleration, and a straight sloped line shows increasing acceleration.
The area under an acceleration-time graph gives the change in velocity

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Katie M

Author: Katie M

Expertise: Curriculum Expert

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

Tim

Reviewer: Tim

Expertise: Content Creator

Timothy graduated with a first class degree in Mathematics and Physics from the University of Warwick. After working as a postgraduate researcher, Timothy has worked as a content creator for various online revision platforms, creating physics resources for a range of levels and exam boards.