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Kinematics

Physics HL Chapter Notes

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KinematicsForces and momentumWork, energy and powerRigid body mechanicsGalilean and special relativity
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1. Distance, Displacement, Speed & Velocity

Kinematics starts by distinguishing between scalar and vector quantities. A scalar has magnitude (size) only, like distance and speed. A vector has both magnitude and direction, like displacement and velocity. Distance is the total length of the path travelled, regardless of direction. Displacement(s) is the object's overall change in position, measured as a straight line from the start point to the end point, including direction. Similarly, speed is a scalar measure of how fast an object is moving (rate of change of distance), while velocity(v) is a vector measure of the rate of change of displacement. Two objects can have the same speed but different velocities if they are moving in different directions.

Average Speed = Total Distance / Time Taken

Average Velocity (v) = Displacement (s) / Time Taken (t)

Key term

Displacement: The change in an object's position, measured as a vector from the starting point to the finishing point.

Common pitfall

Confusing distance with displacement is a classic error. Always consider the start and end points for displacement, and the entire path for distance.

Worked example 14 marks

An athlete runs one full lap of a 400 m circular track in 50 seconds. Calculate:(a) the total distance travelled,(b) her final displacement,(c) her average speed, and(d) her average velocity.

  1. 1

    (a) The distance is the length of the path taken. Since she runs one full lap of the track, the distance is 400 m.

  2. 2

    (b) Displacement is the change in position from start to finish. As she finishes at the same point she started, her displacement is 0 m.

  3. 3

    (c) Average Speed = Total Distance / Time Taken = 400 m / 50 s = 8.0 m/s.

  4. 4

    (d) Average Velocity = Displacement / Time Taken = 0 m / 50 s = 0 m/s.

Recap

  • Distance is a scalar; displacement is a vector.
  • Speed is a scalar; velocity is a vector.
  • Displacement is the straight-line change in position, with direction.
  • Velocity is the rate of change of displacement.
  • If an object returns to its starting point, its displacement is zero.

Quick check

  1. A car travels 50 km North and then 20 km South. What is the distance travelled and what is the displacement?2 marks

2. Acceleration & The SUVAT Equations

Acceleration(a) is the rate of change of velocity. It is a vector quantity, measured in metres per second squared (m/s²). An object is accelerating if it is speeding up, slowing down (decelerating), or changing direction. For motion in a straight line with constant (uniform) acceleration, we can use a set of five special equations known as the SUVAT equations. The variables are: s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time. To solve a problem, you identify three known variables and use the appropriate equation to find the unknown.

v = u + at

s = ut + ½at²

v² = u² + 2as

s = ½(u + v)t

s = vt - ½at²

Key term

Acceleration: The rate of change of an object's velocity with respect to time.

Examiner insight

Examiners reward clear working. Always start by listing your known SUVAT variables, state the equation you are using, and then substitute the values.

Common pitfall

Using the SUVAT equations when acceleration is not constant, or forgetting to use a negative sign for acceleration when an object is slowing down or moving against gravity.

Worked example 14 marks

A car starts from rest and accelerates uniformly at 2.5 m/s² for 6.0 s. Calculate its final velocity and the displacement in this time.

  1. 1
    1. List the known variables: s = ?, u = 0 m/s (from rest), v = ?, a = 2.5 m/s², t = 6.0 s.
  2. 2
    1. To find final velocity (v), choose the equation without s: v = u + at.
  3. 3
    1. Substitute values: v = 0 + (2.5 * 6.0) = 15 m/s.
  4. 4
    1. To find displacement (s), choose an equation with s. Using s = ut + ½at²:
  5. 5
    1. Substitute values: s = (0 * 6.0) + ½(2.5)(6.0)² = 0 + 0.5 * 2.5 * 36 = 45 m.

Worked example 23 marks

A ball is thrown vertically upwards with an initial speed of 20 m/s. Assuming g = 9.81 m/s², what is the maximum height it reaches?

  1. 1
    1. List the variables for the upward journey: s = ? (max height), u = 20 m/s, v = 0 m/s (at max height), a = -9.81 m/s² (acting downwards), t = ?.
  2. 2
    1. To find displacement (s), choose the equation without t: v² = u² + 2as.
  3. 3
    1. Rearrange for s: s = (v² - u²) / 2a.
  4. 4
    1. Substitute values: s = (0² - 20²) / (2 * -9.81) = -400 / -19.62 = 20.4 m.
  5. 5
    1. The maximum height reached is 20.4 m.

Recap

  • Acceleration is the rate of change of velocity.
  • The SUVAT equations only apply when acceleration is constant.
  • For objects in freefall, acceleration 'a' is the acceleration due to gravity, g (approx. 9.81 m/s²).
  • Remember to be consistent with signs for direction (e.g., up is positive, down is negative).
  • At the maximum height of a vertical trajectory, the instantaneous vertical velocity is zero.

Quick check

  1. What are the five variables represented by the letters in SUVAT?2 marks
  2. A train slows down from 30 m/s to 10 m/s in 5 seconds. What is its acceleration?2 marks

3. Analysing Motion with Graphs

Graphs are powerful tools for visualising and analysing motion. The three main types are displacement-time (s-t), velocity-time (v-t), and acceleration-time (a-t) graphs. For a displacement-time graph, the gradient (slope) of the line gives the velocity. A straight line means constant velocity, while a curve means acceleration. For a velocity-time graph, the gradient gives the acceleration, and the area under the graph gives the displacement. A horizontal line means constant velocity (zero acceleration), a straight sloping line means constant acceleration, and a curve means non-uniform acceleration. For an acceleration-time graph, the area under the graph gives the change in velocity.

Velocity = Gradient of a displacement-time graph

Acceleration = Gradient of a velocity-time graph

Displacement = Area under a velocity-time graph

Change in Velocity = Area under an acceleration-time graph

Key term

Gradient: A measure of the steepness of a line on a graph, calculated as the change in the y-axis value divided by the change in the x-axis value (rise/run).

Examiner insight

When asked to find displacement from a velocity-time graph, clearly state that you are calculating the area and show how you are splitting the area into simpler shapes if necessary.

Worked example 14 marks

The velocity-time graph shows the motion of a cyclist.(a) Calculate the acceleration from t=0s to t=10s.(b) Calculate the total displacement of the cyclist after 30s. The graph is a trapezium starting at (0,0), rising to (10,8), flat until (20,8), then falling to (30,0).

  1. 1

    (a) Acceleration is the gradient of the v-t graph. From t=0 to t=10s: Gradient = rise/run = (8 m/s - 0 m/s) / (10 s - 0s) = 8 / 10 = 0.8 m/s².

  2. 2

    (b) Displacement is the area under the v-t graph. The shape is a trapezium. Area = ½(a+b)h, where 'a' and 'b' are the parallel sides and 'h' is the height.

  3. 3

    The top parallel side(b) is from t=10 to t=20, so its length is 10 s. The bottom parallel side(a) is from t=0 to t=30, so its length is 30 s. The height(h) is the maximum velocity, 8 m/s.

  4. 4

    Area = ½(30 + 10) * 8 = ½(40) * 8 = 20 * 8 = 160 m.

  5. 5

    Alternatively, calculate the area of the three parts: Triangle 1: ½*10*8 = 40m. Rectangle: 10*8 = 80m. Triangle 2: ½*10*8 = 40m. Total = 40+80+40 = 160m.

Recap

  • The gradient of an s-t graph is velocity.
  • The gradient of a v-t graph is acceleration.
  • The area under a v-t graph is displacement.
  • A horizontal line on a v-t graph means constant velocity.
  • A straight, sloped line on a v-t graph means constant acceleration.

Quick check

  1. What does a horizontal line on a displacement-time graph represent?1 mark
  2. What quantity is found by calculating the area under an acceleration-time graph?1 mark

4. Projectile Motion

A projectile is any object that is thrown and moves under the influence of gravity alone (air resistance is usually ignored in initial problems). The key to solving projectile motion problems is to treat the horizontal and vertical components of the motion completely independently. The horizontal motion has constant velocity (acceleration is zero). The vertical motion has constant downward acceleration, which is the acceleration due to gravity, g (≈ 9.81 m/s²). You use the standard SUVAT equations for the vertical component and the simple `distance = speed × time` equation for the horizontal component. The time of flight is the link that connects the two components.

Horizontal Motion: aₓ = 0; vₓ = uₓ (constant); sₓ = uₓt

Vertical Motion: aᵧ = -g (≈ -9.81 m/s²); vᵧ = uᵧ + aᵧt; sᵧ = uᵧt + ½aᵧt²; vᵧ² = uᵧ² + 2aᵧsᵧ

Key term

Projectile: An object that is given an initial velocity and then moves freely under the influence of gravity.

Examiner insight

For questions about maximum height, the key piece of information is that the vertical velocity (vᵧ) is momentarily zero at the peak.

Common pitfall

Forgetting to resolve the initial velocity into horizontal and vertical components, or mixing the components in calculations.

Worked example 15 marks

A golf ball is hit with an initial velocity of 40 m/s at an angle of 30° to the horizontal. Calculate:(a) the time of flight, and(b) the horizontal range. (Use g = 9.81 m/s²).

  1. 1
    1. Resolve the initial velocity into components. uₓ = u cos(θ) = 40 * cos(30°) = 34.64 m/s. uᵧ = u sin(θ) = 40 * sin(30°) = 20 m/s.
  2. 2
    1. (a) To find the time of flight, consider the vertical motion until it returns to the ground (sᵧ = 0). Use sᵧ = uᵧt + ½aᵧt².
  3. 3
    1. 0 = 20t + ½(-9.81)t² = 20t - 4.905t². Factorise: t(20 - 4.905t) = 0.
  4. 4
    1. The solutions are t=0 (the start) and 20 - 4.905t = 0. So, t = 20 / 4.905 = 4.08 s. The time of flight is 4.08 s.
  5. 5
    1. (b) To find the horizontal range (sₓ), use the horizontal motion equation: sₓ = uₓt.
  6. 6
    1. Substitute values: sₓ = 34.64 m/s * 4.08 s = 141 m.

Recap

  • Split projectile motion into independent horizontal and vertical components.
  • Horizontal acceleration is zero (constant horizontal velocity).
  • Vertical acceleration is constant and downwards (g ≈ 9.81 m/s²).
  • Time is the common variable linking horizontal and vertical motion.
  • At the peak of the trajectory, the vertical component of velocity is zero.

Quick check

  1. For a projectile launched at an angle, what is its acceleration at the highest point of its path?1 mark

5. Air Resistance and Terminal Velocity

In real-world situations, a fluid resistance force, such as air resistance, opposes the motion of an object. This force is not constant; it increases as the object's speed increases. For a projectile, air resistance reduces its maximum height and range, and makes the flight path asymmetrical (the descent is steeper than the ascent). For a falling object, like a skydiver, air resistance increases as they speed up. Eventually, the upward force of air resistance becomes equal in magnitude to the downward force of weight. At this point, the net force on the object is zero, so its acceleration becomes zero. The object stops accelerating and falls at a constant maximum speed, known as terminal velocity.

Key term

Terminal Velocity: The constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling equals the force of gravity.

Fun fact

A cat has a much lower terminal velocity (about 100 km/h) than a human (about 200 km/h). This, combined with their flexibility, is why they can often survive falls from great heights.

Worked example 14 marks

A skydiver jumps from a plane. Describe and explain the change in their velocity and acceleration from the moment they jump until they reach terminal velocity.

  1. 1
    1. Initially, just after jumping, the only significant force is weight. Speed is low, so air resistance is negligible. The skydiver accelerates downwards at approximately g (9.81 m/s²).
  2. 2
    1. As velocity increases, the upward force of air resistance increases. This opposes the weight, so the net downward force decreases.
  3. 3
    1. Since F=ma and the net force is decreasing, the downward acceleration decreases. The skydiver is still speeding up, but at a slower rate.
  4. 4
    1. Eventually, the skydiver's speed is high enough that the force of air resistance becomes equal in magnitude to their weight. The net force is now zero.
  5. 5
    1. With zero net force, acceleration is zero. The skydiver's velocity becomes constant. This constant velocity is the terminal velocity.

Recap

  • Air resistance is a drag force that opposes motion and increases with speed.
  • Air resistance causes the range and maximum height of a projectile to decrease.
  • Terminal velocity is reached when the force of air resistance equals the weight of the falling object.
  • At terminal velocity, the net force is zero and acceleration is zero.
  • The velocity-time graph for an object reaching terminal velocity is a curve that flattens to a horizontal line.

Quick check

  1. What is the net force on an object falling at its terminal velocity?1 mark

End-of-chapter exercise

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

  1. A student walks 80 m east and then 50 m west. The entire journey takes 100 seconds. Calculate the student's (a) total distance travelled, (b) final displacement, (c) average speed, and (d) average velocity.4 marks
  2. A rocket accelerates from rest to a speed of 150 m/s in 4.0 s. What is its average acceleration?2 marks
  3. A stone is dropped from a 50 m high cliff. How long does it take to hit the water below? (Use g = 9.81 m/s²).3 marks
  4. Explain the difference between a scalar and a vector quantity, giving one example of each from kinematics.3 marks
  5. A velocity-time graph for a car's journey is a triangle with vertices at (0,0), (5,10), and (10,0). Calculate (a) the acceleration during the first 5 seconds, and (b) the total displacement for the 10-second journey.4 marks
  6. A cannonball is fired horizontally at 80 m/s from the top of a 120 m high cliff. How far from the base of the cliff does the cannonball land?4 marks
  7. Describe the motion of an object represented by a horizontal line on (a) a displacement-time graph, and (b) a velocity-time graph.2 marks
  8. A skydiver of mass 75 kg reaches a terminal velocity of 50 m/s. What is the magnitude of the air resistance force acting on them at this speed? (Use g = 9.81 m/s²).3 marks
  9. A ball is thrown from ground level with an initial velocity of 28 m/s at an angle of 40° above the horizontal. What is the maximum height reached by the ball?4 marks
  10. A car accelerates from rest at 2.0 m/s² for 5.0 s. It then travels at a constant velocity for 10 s before braking uniformly to a stop over a distance of 50 m. Calculate the total distance travelled by the car.6 marks

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