Cambridge AS & A Level9702

Kinematics

Physics 9702 Chapter Notes

What this chapter covers

Kinematics - Equations of motion
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In this chapter

  1. 1Scalars and Vectors
  2. 2Displacement-Time Graphs
  3. 3Velocity-Time Graphs and Acceleration
  4. 4Equations of Uniformly Accelerated Motion (SUVAT)
  5. 5Free Fall and Acceleration 'g'
  6. 6Introduction to Projectile Motion

1. Scalars and Vectors

In physics, quantities can be either scalars or vectors. A scalar quantity has only magnitude (a size or amount). Examples include distance (e.g., 500 m), speed (e.g., 25 m/s), mass, and time. A vector quantity has both magnitude and direction. Examples include displacement (e.g., 500 m due East), velocity (e.g., 25 m/s due East), acceleration, and force. Two cars can have the same speed but different velocities if they are travelling in different directions. Distance is the total path length covered, while displacement is the straight-line distance from the start point to the end point, in a specific direction.

Average Speed = Total Distance / Total Time

Average Velocity = Total Displacement / Total Time

Key term

Displacement: The straight-line distance of an object from its starting point in a specific direction.

Examiner insight

Examiners frequently test the distinction between distance (a scalar) and displacement (a vector). Always consider direction when dealing with vector quantities.

Common pitfall

Confusing distance and displacement. Displacement can be zero even when a large distance has been travelled, for example, after one lap of a circular track.

Worked example 14 marks

A runner jogs 400 m East, then turns around and jogs 100 m West. The entire journey takes 100 seconds. Calculate:(a) the total distance travelled,(b) the runner's displacement,(c) the average speed, and(d) the average velocity.

  1. 1

    Step 1 (a): Find the total distance. This is the sum of the path lengths. Distance = 400 m + 100 m = 500 m.

  2. 2

    Step 2 (b): Find the displacement. This is the final position relative to the start. Let East be the positive direction. Displacement = (+400m) + (-100m) = +300 m. So, the displacement is 300 m East.

  3. 3

    Step 3 (c): Calculate the average speed. Average Speed = Total Distance / Total Time = 500 m / 100 s = 5.0 m s⁻¹.

  4. 4

    Step 4 (d): Calculate the average velocity. Average Velocity = Total Displacement / Total Time = 300 m East / 100 s = 3.0 m s⁻¹ East.

Recap

  • A scalar quantity has magnitude only.
  • A vector quantity has both magnitude and direction.
  • Distance is a scalar, while displacement is a vector.
  • Speed is a scalar, while velocity is a vector.
  • Displacement is the change in position (Δs) in a specific direction.

Quick check

  1. Is temperature a scalar or a vector quantity?1 mark
  2. An object travels one full lap of a 400 m track. What is its displacement?1 mark

2. Displacement-Time Graphs

A displacement-time graph plots an object's displacement from a fixed origin against time. The gradient (steepness) of the line on this graph tells you the object's velocity. A straight line indicates constant velocity. A horizontal line means the object is stationary (velocity = 0). A curved line indicates that the velocity is changing, which means the object is accelerating. A steeper line means a greater velocity.

Velocity = Gradient = Change in Displacement / Change in Time = Δs / Δt

Key term

Gradient: The steepness of a line on a graph, which represents the rate of change of the y-axis quantity with respect to the x-axis quantity.

Examiner insight

Candidates who can accurately calculate the gradient of a tangent to a curve to find the instantaneous velocity at a specific point will access the highest marks.

Common pitfall

Confusing a displacement-time graph with a velocity-time graph. A horizontal line on a displacement-time graph means the object is stationary, not moving at a constant velocity.

Worked example 13 marks

The displacement-time graph shows the motion of a cyclist.(a) Describe the motion in the first 10 s.(b) Calculate the velocity between 10 s and 30 s.(c) What is the total displacement after 40 s?

  1. 1

    Step 1 (a): In the first 10 s, the graph is a straight line sloping upwards, starting from the origin. This represents constant positive velocity.

  2. 2

    Step 2 (b): Velocity is the gradient. Between t=10s and t=30s, the displacement changes from 50m to 50m. Gradient = Δs / Δt = (50 - 50) / (30 - 10) = 0 / 20 = 0 m s⁻¹. The cyclist is stationary.

  3. 3

    Step 3 (c): To find the total displacement after 40 s, we simply read the value from the y-axis at t=40 s. The graph shows the line returning to s=0 at t=40s. The total displacement is 0 m.

Recap

  • The gradient of a displacement-time graph is equal to velocity.
  • A horizontal line on a displacement-time graph means the object is stationary.
  • A straight, sloped line represents constant velocity.
  • A curved line represents acceleration (changing velocity).
  • The y-intercept represents the initial displacement.

Quick check

  1. What does a straight, horizontal line on a displacement-time graph represent?1 mark

3. Velocity-Time Graphs and Acceleration

A velocity-time graph plots velocity against time. It provides a wealth of information. The gradient of the line represents acceleration (rate of change of velocity). The area under the graph represents the displacement. A horizontal line means constant velocity (zero acceleration). A straight, sloped line indicates constant acceleration. A curved line means the acceleration is changing (non-uniform acceleration).

Acceleration = Gradient = Change in Velocity / Time Taken = (v - u) / t

Displacement = Area under velocity-time graph

Key term

Acceleration: The rate of change of velocity, a vector quantity measured in metres per second squared (m s⁻²).

Examiner insight

Marks are often awarded for correctly calculating the area of a trapezium to find displacement in a single step, which is more efficient than splitting it into a rectangle and a triangle.

Common pitfall

Calculating the area under the graph to find the distance, but forgetting to account for negative displacement when the velocity is negative (i.e., below the time axis).

Worked example 14 marks

A car starts from rest and its motion is shown on the velocity-time graph. Calculate:(a) the acceleration in the first 4 seconds, and(b) the total displacement after 10 seconds.

  1. 1

    Step 1 (a): Acceleration is the gradient of the v-t graph. For the first 4 seconds: a = Δv / Δt = (8 m s⁻¹ - 0 m s⁻¹) / (4 s - 0s) = 8 / 4 = 2 m s⁻².

  2. 2

    Step 2 (b): Displacement is the area under the graph. We can split the area into a triangle (0-4s) and a rectangle (4-10s).

  3. 3

    Step 3: Area of triangle = ½ × base × height = ½ × 4 s × 8 m s⁻¹ = 16 m.

  4. 4

    Step 4: Area of rectangle = width × height = (10 - 4) s × 8 m s⁻¹ = 6 s × 8 m s⁻¹ = 48 m.

  5. 5

    Step 5: Total displacement = Area of triangle + Area of rectangle = 16 m + 48 m = 64 m.

Recap

  • The gradient of a velocity-time graph is equal to acceleration.
  • The area under a velocity-time graph is equal to displacement.
  • A horizontal line represents constant velocity (zero acceleration).
  • A straight, sloped line represents constant (uniform) acceleration.
  • Area below the time axis represents displacement in the negative direction.

Quick check

  1. How do you determine the total displacement from a velocity-time graph?1 mark

4. Equations of Uniformly Accelerated Motion (SUVAT)

For any object moving with constant (uniform) acceleration, its motion can be described by a set of four equations, often called the SUVAT equations. These equations relate five variables: s (displacement), u (initial velocity), v (final velocity), a (acceleration), and t (time). If you know any three of these variables, you can use these equations to find the other two. Remember: these equations ONLY apply when acceleration is constant.

v = u + at

s = ut + ½at²

v² = u² + 2as

s = ½(u + v)t

Key term

Uniform Acceleration: A constant rate of change of velocity, meaning the velocity changes by the same amount in each equal time interval.

Examiner insight

A systematic approach is rewarded. Students should list the known SUVAT variables, identify the unknown, and then select the appropriate equation that links them to avoid careless mistakes.

Common pitfall

Using the SUVAT equations in situations where acceleration is not constant, for example, for a whole journey that involves speeding up and then slowing down.

Worked example 13 marks

A car travelling at 10 m s⁻¹ accelerates uniformly at 2.0 m s⁻² for 5.0 s. Calculate its final velocity.

  1. 1

    Step 1: List the known variables. u = 10 m s⁻¹, a = 2.0 m s⁻², t = 5.0 s.

  2. 2

    Step 2: Identify the unknown variable. We want to find v.

  3. 3

    Step 3: Select the equation that links u, a, t, and v. This is v = u + at.

  4. 4

    Step 4: Substitute the values and solve. v = 10 + (2.0 × 5.0) = 10 + 10 = 20 m s⁻¹.

Worked example 23 marks

A train starts from rest and accelerates uniformly. It travels a distance of 200 m in 20 s. What is its acceleration?

  1. 1

    Step 1: List the known variables. s = 200 m, t = 20 s, u = 0 m s⁻¹ (starts from rest).

  2. 2

    Step 2: Identify the unknown variable. We want to find a.

  3. 3

    Step 3: Select the equation that links s, u, t, and a. This is s = ut + ½at².

  4. 4

    Step 4: Substitute the values and solve for a. 200 = (0 × 20) + ½ × a × (20)².

  5. 5

    Step 5: Simplify the equation. 200 = 0 + ½ × a × 400. So, 200 = 200a.

  6. 6

    Step 6: Calculate a. a = 200 / 200 = 1.0 m s⁻².

Recap

  • The SUVAT equations only apply for motion with constant acceleration.
  • Always list your known variables (s, u, v, a, t) before starting.
  • Choose the equation that contains your known variables and the one you want to find.
  • Remember that 'from rest' means u = 0.
  • Deceleration is negative acceleration.

Quick check

  1. Which SUVAT equation should you use if you do not know the time 't'?1 mark

5. Free Fall and Acceleration 'g'

An object in free fall is moving solely under the influence of gravity. We assume that air resistance is negligible. Near the Earth's surface, all objects in free fall have the same constant, downward acceleration, known as the acceleration of free fall, 'g'. The accepted value is g ≈ 9.81 m s⁻². Since this acceleration is constant, we can use the SUVAT equations for all free fall problems. It's crucial to be consistent with signs; a common convention is to take the upward direction as positive, which means acceleration 'a' is always -g (-9.81 m s⁻²).

a = g ≈ 9.81 m s⁻² (downwards)

Key term

Free Fall: The motion of an object under the influence of gravity alone, with no other forces such as air resistance acting on it.

Examiner insight

Be careful with signs. A common convention is to take 'up' as the positive direction, meaning displacement and velocity are positive upwards, but acceleration 'g' is always negative (-9.81 m s⁻²).

Common pitfall

Thinking that the acceleration of an object at the highest point of its trajectory is zero. Its velocity is momentarily zero, but its acceleration is still g downwards.

Worked example 15 marks

A stone is dropped from rest from a 50 m high cliff. Ignoring air resistance, calculate(a) the time it takes to hit the ground and(b) its velocity just before impact. (Use g = 9.81 m s⁻²).

  1. 1

    Step 1: List the variables. Let's take downwards as the positive direction. s = +50 m, u = 0 m s⁻¹, a = +g = +9.81 m s⁻².

  2. 2

    Step 2 (a): Find time 't'. Use s = ut + ½at². 50 = (0)t + ½(9.81)t². 50 = 4.905t².

  3. 3

    Step 3: Solve for t. t² = 50 / 4.905 = 10.19. So, t = √10.19 = 3.19 s.

  4. 4

    Step 4 (b): Find final velocity 'v'. Use v² = u² + 2as. v² = 0² + 2(9.81)(50).

  5. 5

    Step 5: Solve for v. v² = 981. So, v = √981 = 31.3 m s⁻¹ (downwards).

Recap

  • In free fall, the only force acting is gravity, so acceleration is constant (g ≈ 9.81 m s⁻²).
  • Air resistance is assumed to be negligible in free fall problems.
  • The SUVAT equations can be used for vertical motion by setting a = g.
  • Be consistent with your choice of positive direction (up or down).
  • At the highest point of a trajectory, the vertical velocity is momentarily zero, but acceleration is still g.

Quick check

  1. A ball is thrown vertically upwards. What is its acceleration at its highest point?1 mark

6. Introduction to Projectile Motion

A projectile is any object that is thrown and moves through the air under the influence of gravity alone (air resistance is ignored). The key to solving projectile problems is to realise that the motion can be split into two independent components: horizontal motion and vertical motion. The horizontal velocity is constant because there is no horizontal acceleration. The vertical velocity changes because of the constant downward acceleration due to gravity, g. Time is the one quantity that is common to both the horizontal and vertical motions.

Horizontal motion: a_x = 0, v_x = constant, s_x = v_x * t

Vertical motion: a_y = -g, use SUVAT equations

v_x = v cos(θ) (horizontal component of initial velocity)

v_y = v sin(θ) (vertical component of initial velocity)

Key term

Projectile: An object that is thrown or projected into the air and moves under the influence of gravity alone.

Examiner insight

Clear working that separates the horizontal and vertical components of motion is essential. Use two separate columns or sections in your calculations to avoid confusion.

Common pitfall

Mixing up horizontal and vertical components in calculations. For example, using the vertical acceleration 'g' in a horizontal motion calculation.

Fun fact

The curved path, or trajectory, of a projectile is a parabola. This mathematical shape is seen everywhere from the arc of a water fountain to the path of a basketball shot.

Worked example 15 marks

A ball is kicked horizontally with a velocity of 15 m s⁻¹ from the top of a cliff that is 45 m high. How far from the base of the cliff does the ball land? (Use g = 9.81 m s⁻²).

  1. 1

    Step 1: Separate the motion into horizontal and vertical components.

  2. 2

    Step 2: Analyse the vertical motion to find the time of flight. s_y = 45 m, u_y = 0 m s⁻¹, a_y = 9.81 m s⁻². We need to find t.

  3. 3

    Step 3: Use s = ut + ½at². 45 = (0)t + ½(9.81)t². 45 = 4.905t².

  4. 4

    Step 4: Solve for t. t² = 45 / 4.905 = 9.17. So, t = √9.17 = 3.03 s. This is the time the ball is in the air.

  5. 5

    Step 5: Analyse the horizontal motion to find the range. v_x = 15 m s⁻¹ (constant), t = 3.03 s.

  6. 6

    Step 6: Use distance = speed × time. Horizontal distance (range) s_x = v_x × t = 15 m s⁻¹ × 3.03 s = 45.5 m.

Recap

  • Projectile motion is analysed by splitting it into independent horizontal and vertical components.
  • The horizontal component of velocity is constant (acceleration is zero).
  • The vertical component of motion is subject to constant downward acceleration, g.
  • The time of flight is the same for both horizontal and vertical motions.
  • Resolve the initial velocity into horizontal (v cos θ) and vertical (v sin θ) components if launched at an angle.

Quick check

  1. For a projectile launched in the absence of air resistance, which component of its velocity remains constant?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A car accelerates uniformly from 5.0 m s⁻¹ to 25 m s⁻¹ in 8.0 s. Calculate the acceleration of the car and the distance it travels in this time.4 marks
  2. Define velocity and explain why it is a vector quantity.2 marks
  3. The velocity-time graph for a lift is shown. Calculate the total distance travelled by the lift during the 12 seconds.3 marks
  4. A ball is thrown vertically upwards with an initial speed of 20 m s⁻¹. Ignoring air resistance, what is the maximum height it reaches? (Use g = 9.81 m s⁻²).3 marks
  5. An aircraft must reach a speed of 60 m s⁻¹ for take-off. If the runway is 900 m long, what is the minimum uniform acceleration required, assuming the aircraft starts from rest?3 marks
  6. A stone is dropped down a well and the splash is heard 3.0 s later. If the depth of the well is 44.1 m, calculate the speed of sound in air. (Use g = 9.81 m s⁻²).5 marks
  7. Describe an experiment to determine the acceleration of free fall, g, using a falling object and electronic timers. State the measurements you would take and how you would use them to find g.6 marks
  8. A golf ball is struck with a velocity of 40 m s⁻¹ at an angle of 30° to the horizontal. Calculate (a) the time of flight and (b) the horizontal range of the ball. (Ignore air resistance, use g = 9.81 m s⁻²).6 marks
  9. A cyclist is travelling at a constant velocity of 4.0 m s⁻¹. They pass a stationary runner who immediately starts to accelerate at 0.5 m s⁻² in the same direction. How long does it take for the runner to catch the cyclist?4 marks
  10. From the top of a 70 m high building, a ball is thrown upwards with a speed of 20 m s⁻¹. How long does it take for the ball to reach the ground? (Use g = 9.81 m s⁻²).4 marks

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