Mechanics

Kinematics: Motion, Forces and Energy

Year 12 · Year 13

✓Interpret and sketch displacement-time, velocity-time, and acceleration-time graphs, understanding the relationships between them.
✓Apply the SUVAT equations to solve problems involving particles moving with constant acceleration in a straight line.
✓Use differentiation and integration to solve problems involving variable acceleration in one dimension.
✓Analyse and solve problems involving projectile motion under gravity, treating horizontal and vertical components independently.

Key concepts

Motion Graphs

Motion graphs visually represent the movement of a particle. The key types are displacement-time (s-t), velocity-time (v-t), and acceleration-time (a-t) graphs. The gradient of an s-t graph gives velocity, and the gradient of a v-t graph gives acceleration. The area under a v-t graph gives displacement, and the area under an a-t graph gives the change in velocity.

SUVAT Equations (Constant Acceleration)

These are a set of five equations used to solve problems involving motion in a straight line with constant acceleration. 's' is displacement, 'u' is initial velocity, 'v' is final velocity, 'a' is constant acceleration, and 't' is time.

v = u + at, s = ut + ½at², s = vt - ½at², s = ½(u + v)t, v² = u² + 2as

Variable Acceleration (Calculus)

When acceleration is not constant, calculus is used to relate displacement, velocity, and acceleration. Velocity is the derivative of displacement with respect to time, and acceleration is the derivative of velocity with respect to time. Conversely, integrating acceleration gives velocity, and integrating velocity gives displacement.

v = ds/dt, a = dv/dt = d²s/dt², s = ∫v dt, v = ∫a dt

Projectile Motion (A-Level)

Projectile motion describes the path of an object thrown into the air, subject only to gravity. Key assumptions include negligible air resistance and constant acceleration due to gravity (g = 9.8 m/s²). The horizontal and vertical components of motion are treated independently. Horizontal motion has constant velocity, while vertical motion has constant acceleration 'g' downwards.

Horizontal: s_x = u_x t, a_x = 0. Vertical: s_y = u_y t + ½gt², v_y = u_y + gt, v_y² = u_y² + 2gs_y. Initial components for projection at angle α: u_x = U cos α, u_y = U sin α

Key facts to remember

1Displacement is a vector quantity (magnitude and direction) representing the change in position from a fixed origin.
2Velocity is the rate of change of displacement, and acceleration is the rate of change of velocity; both are vector quantities.
3For constant acceleration, the five SUVAT equations are essential tools for solving motion problems.
4The gradient of a displacement-time graph gives velocity, and the gradient of a velocity-time graph gives acceleration.
5The area under a velocity-time graph gives displacement, and the area under an acceleration-time graph gives the change in velocity.
6For variable acceleration, calculus provides the relationships: v = ds/dt, a = dv/dt, s = ∫v dt, v = ∫a dt.
7In projectile motion, horizontal and vertical components of motion are independent. Horizontal acceleration is zero, and vertical acceleration is g (9.8 m/s² or 9.81 m/s²) downwards.
8Always define a consistent positive direction for vector quantities (displacement, velocity, acceleration) at the start of a problem.

Worked examples

Example 1

A car accelerates uniformly from rest to 20 m/s in 8 seconds. Calculate the distance travelled during this time.

IIdentify known variables: initial velocity u = 0 m/s (from rest), final velocity v = 20 m/s, time t = 8 s.

IIIdentify unknown variable: displacement s.

IIIChoose the appropriate SUVAT equation that relates s, u, v, and t: s = ½(u + v)t.

IVSubstitute the known values into the equation: s = ½(0 + 20) × 8.

VCalculate the result: s = ½(20) × 8 = 10 × 8 = 80.

Answer

The distance travelled is 80 m.

Example 2

A particle P moves along the x-axis. At time t seconds, its displacement s metres from the origin O is given by s = t³ - 6t² + 9t. a) Find an expression for the velocity of P at time t. b) Find the times when P is instantaneously at rest. c) Find an expression for the acceleration of P at time t.

Ia) To find velocity, differentiate displacement with respect to time: v = ds/dt.

II v = d/dt (t³ - 6t² + 9t) = 3t² - 12t + 9.

IIIb) P is instantaneously at rest when its velocity is zero. Set v = 0.

IV 3t² - 12t + 9 = 0.

V Divide by 3: t² - 4t + 3 = 0.

VI Factorise the quadratic equation: (t - 1)(t - 3) = 0.

VII Solve for t: t = 1 or t = 3.

VIIIc) To find acceleration, differentiate velocity with respect to time: a = dv/dt.

9 a = d/dt (3t² - 12t + 9) = 6t - 12.

Answer

a) v = 3t² - 12t + 9 m/s. b) P is at rest when t = 1 second and t = 3 seconds. c) a = 6t - 12 m/s².

Remember to include the units for each quantity.

Example 3

A stone is projected from a point O on horizontal ground with speed 15 m/s at an angle of 30° above the horizontal. Take g = 9.8 m/s². a) Find the time of flight of the stone. b) Find the horizontal range of the stone.

I1. Resolve the initial velocity (U = 15 m/s, α = 30°) into horizontal and vertical components:

II u_x = U cos α = 15 cos 30° = 15 × (√3 / 2) = 7.5√3 m/s.

III u_y = U sin α = 15 sin 30° = 15 × (1 / 2) = 7.5 m/s.

IV2. Consider vertical motion (taking upwards as positive, so acceleration a = -g = -9.8 m/s²):

V Initial vertical velocity u_y = 7.5 m/s.

VI Vertical displacement s_y = 0 (when the stone returns to ground level).

VII Acceleration a = -9.8 m/s².

VIII Time t = ?

9a) To find the time of flight, use the SUVAT equation s_y = u_y t + ½at²:

10 0 = 7.5t + ½(-9.8)t².

11 0 = 7.5t - 4.9t².

12 Factorise t: t(7.5 - 4.9t) = 0.

13 This gives t = 0 (initial projection) or 7.5 - 4.9t = 0.

14 4.9t = 7.5.

15 t = 7.5 / 4.9 ≈ 1.5306 seconds.

16b) To find the horizontal range, use the horizontal motion equation s_x = u_x t (since horizontal acceleration is 0):

17 s_x = (7.5√3) × (7.5 / 4.9).

18 s_x ≈ 12.990 × 1.5306 ≈ 19.90 m.

Answer

a) The time of flight is approximately 1.53 seconds (3 s.f.). b) The horizontal range of the stone is approximately 19.9 metres (3 s.f.).

Always define your positive direction for vertical motion. Here, upwards was positive, so gravity was negative.

Common mistakes

✗Confusing scalar quantities (distance, speed) with vector quantities (displacement, velocity) and using them interchangeably.
✗Incorrectly applying SUVAT equations when the acceleration is not constant, or when the motion is not in a straight line.
✗Making sign errors with acceleration due to gravity (g) in vertical motion problems, especially when defining upwards as the positive direction.
✗Forgetting to include the constant of integration when using calculus and failing to use initial conditions to determine its value.
✗Mixing units (e.g., km/h with m/s, or minutes with seconds) without performing the necessary conversions.

Exam tips

★Always draw a clear diagram for projectile motion problems, labelling initial velocity components, angles, and the path. This helps visualise the problem.
★For SUVAT problems, list the known variables (s, u, v, a, t) and the unknown you need to find. This systematic approach helps in selecting the correct formula.
★Be consistent with your choice of positive direction for all vector quantities throughout a problem (e.g., upwards positive, rightwards positive). State your convention clearly.
★Show all steps in your working, especially for calculus problems and projectile motion, to gain full method marks, even if your final answer has a minor arithmetic error.

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