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Mechanics

Kinematics

5th Year · 6th Year (Leaving Cert)

  • ✓By the end of this lesson students will be able to define and distinguish between displacement, velocity, and acceleration.
  • ✓By the end of this lesson students will be able to apply the equations of motion for objects moving with constant acceleration in a straight line.
  • ✓By the end of this lesson students will be able to interpret and draw displacement-time and velocity-time graphs.
  • ✓By the end of this lesson students will be able to solve problems involving projectile motion, including calculating range, maximum height, and time of flight (Higher Level).

Key concepts

Displacement (s)

The shortest distance from the starting point to the finishing point in a specified direction. It is a vector quantity, meaning it has both magnitude and direction. Its SI unit is the metre (m).

Velocity (v or u)

The rate of change of displacement. It is a vector quantity. 'u' typically denotes initial velocity, and 'v' denotes final velocity. Its SI unit is metres per second (m s⁻¹).

v = Δs / Δt
Acceleration (a)

The rate of change of velocity. It is a vector quantity. A positive acceleration means velocity is increasing in the positive direction, while a negative acceleration (deceleration) means velocity is decreasing in the positive direction or increasing in the negative direction. Its SI unit is metres per second squared (m s⁻²).

a = Δv / Δt
Equations of Motion (SUVAT Equations)

These equations describe the motion of objects moving with constant acceleration in a straight line. They relate displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t).

v = u + at s = ut + ½at² v² = u² + 2as s = (u+v)/2 * t
Displacement-time Graphs

A graph showing how an object's displacement changes over time. The gradient (slope) of a displacement-time graph represents the object's velocity. A straight line indicates constant velocity, while a curved line indicates changing velocity (acceleration).

Velocity-time Graphs

A graph showing how an object's velocity changes over time. The gradient (slope) of a velocity-time graph represents the object's acceleration. The area under the graph represents the object's displacement.

Projectile Motion (Higher Level)

The motion of an object projected into the air, subject only to the acceleration due to gravity (ignoring air resistance). The motion is analysed by resolving the initial velocity into independent horizontal and vertical components. The horizontal component of velocity remains constant, while the vertical component of motion is subject to constant acceleration 'g' (approximately 9.8 m s⁻² downwards).

Horizontal motion: s_x = u_x * t, v_x = u_x (constant) Vertical motion: s_y = u_y * t + ½gt², v_y = u_y + gt, v_y² = u_y² + 2gs_y (where 'g' is acceleration due to gravity, typically taken as negative if upwards is positive)

Key facts to remember

  • 1Displacement, velocity, and acceleration are vector quantities, possessing both magnitude and direction.
  • 2The equations of motion (SUVAT equations) are only applicable for objects moving with constant acceleration in a straight line.
  • 3On a displacement-time graph, the gradient represents velocity.
  • 4On a velocity-time graph, the gradient represents acceleration, and the area under the graph represents displacement.
  • 5In projectile motion (ignoring air resistance), the horizontal component of velocity remains constant, while the vertical component of motion is subject to constant acceleration due to gravity (g ≈ 9.8 m s⁻² downwards).
  • 6The acceleration due to gravity, g, is always directed downwards.

Worked examples

Example 1

A car accelerates uniformly from rest to a velocity of 25 m s⁻¹ in 10 seconds. Calculate (i) its acceleration and (ii) the distance travelled during this time.

IList the knowns: u = 0 m s⁻¹ (from rest), v = 25 m s⁻¹, t = 10 s.
IITo find acceleration (a), use the equation v = u + at.
IIISubstitute values: 25 = 0 + a(10).
IVSolve for a: 10a = 25 => a = 25/10 = 2.5 m s⁻².
VTo find the distance travelled (s), use the equation s = ut + ½at².
VISubstitute values: s = (0)(10) + ½(2.5)(10)².
VIICalculate: s = 0 + ½(2.5)(100) = ½(250) = 125 m.

Answer

(i) Acceleration = 2.5 m s⁻² (ii) Distance travelled = 125 m

Alternatively, for distance, you could use v² = u² + 2as: 25² = 0² + 2(2.5)s => 625 = 5s => s = 125 m.

Example 2

A velocity-time graph shows an object starting from rest, accelerating uniformly to 12 m s⁻¹ in 3 seconds, then moving at a constant velocity for 5 seconds, and finally decelerating uniformly to rest in 2 seconds. Calculate (i) the acceleration during each phase and (ii) the total displacement of the object.

IPhase 1 (Acceleration): u = 0 m s⁻¹, v = 12 m s⁻¹, t = 3 s.
IIAcceleration (a₁) = (v - u) / t = (12 - 0) / 3 = 4 m s⁻².
IIIDisplacement (s₁) = Area of triangle = ½ * base * height = ½ * 3 * 12 = 18 m.
IVPhase 2 (Constant Velocity): u = 12 m s⁻¹, v = 12 m s⁻¹, t = 5 s.
VAcceleration (a₂) = (v - u) / t = (12 - 12) / 5 = 0 m s⁻².
VIDisplacement (s₂) = Area of rectangle = base * height = 5 * 12 = 60 m.
VIIPhase 3 (Deceleration): u = 12 m s⁻¹, v = 0 m s⁻¹, t = 2 s.
VIIIAcceleration (a₃) = (v - u) / t = (0 - 12) / 2 = -6 m s⁻².
9Displacement (s₃) = Area of triangle = ½ * base * height = ½ * 2 * 12 = 12 m.
10Total displacement = s₁ + s₂ + s₃ = 18 + 60 + 12 = 90 m.

Answer

(i) Phase 1 acceleration = 4 m s⁻²; Phase 2 acceleration = 0 m s⁻²; Phase 3 acceleration = -6 m s⁻². (ii) Total displacement = 90 m.

A negative acceleration indicates deceleration in the direction of motion.

Example 3

(Higher Level) A projectile is fired from the ground with an initial velocity of 40 m s⁻¹ at an angle of 30° above the horizontal. Take g = 9.8 m s⁻². Calculate: (i) The time taken to reach its maximum height. (ii) The maximum height reached. (iii) The horizontal range of the projectile.

IResolve the initial velocity (u = 40 m s⁻¹) into horizontal (u_x) and vertical (u_y) components:
IIu_x = u cos θ = 40 cos 30° = 40 * (√3 / 2) ≈ 34.64 m s⁻¹.
IIIu_y = u sin θ = 40 sin 30° = 40 * (1 / 2) = 20 m s⁻¹.
IVFor vertical motion, take upwards as positive, so acceleration a = -g = -9.8 m s⁻².
V(i) To find time to maximum height (t_max), at maximum height, the vertical velocity (v_y) is 0.
VIUse v_y = u_y + at_max => 0 = 20 + (-9.8)t_max.
VII9.8t_max = 20 => t_max = 20 / 9.8 ≈ 2.04 s.
VIII(ii) To find maximum height (s_y_max), use v_y² = u_y² + 2as_y_max.
90² = 20² + 2(-9.8)s_y_max => 0 = 400 - 19.6s_y_max.
1019.6s_y_max = 400 => s_y_max = 400 / 19.6 ≈ 20.41 m.
11(iii) To find the horizontal range, first find the total time of flight (T). The total time of flight is twice the time to maximum height (assuming it lands at the same level it was fired from).
12T = 2 * t_max = 2 * 2.04 = 4.08 s.
13Alternatively, use s_y = u_y T + ½aT² with s_y = 0 (back to ground level): 0 = 20T + ½(-9.8)T² => 0 = 20T - 4.9T².
14Factorise: T(20 - 4.9T) = 0. So T = 0 (start) or 20 - 4.9T = 0 => 4.9T = 20 => T = 20 / 4.9 ≈ 4.08 s.
15Horizontal range (s_x) = u_x * T (since horizontal velocity is constant).
16s_x = 34.64 * 4.08 ≈ 141.24 m.

Answer

(i) Time to maximum height ≈ 2.04 s (ii) Maximum height reached ≈ 20.41 m (iii) Horizontal range ≈ 141.24 m

Always resolve initial velocity into components for projectile motion. Be consistent with your chosen positive direction for vertical motion (e.g., upwards positive, so g is negative).

Common mistakes

  • ✗Confusing scalar quantities (distance, speed) with vector quantities (displacement, velocity).
  • ✗Applying the equations of motion when acceleration is not constant or when motion is not in a straight line.
  • ✗Incorrectly assigning positive or negative signs to vector quantities like displacement, velocity, and acceleration (especially 'g' in vertical motion).
  • ✗Forgetting to resolve initial velocity into its horizontal and vertical components in projectile motion problems.
  • ✗Using the wrong formula or mixing up variables (e.g., using 's' for speed instead of displacement).

Exam tips

  • ★Always begin by listing all known quantities (u, v, a, s, t) and clearly identify the unknown quantity you need to find. This helps in selecting the correct equation.
  • ★Choose a consistent positive direction for your calculations (e.g., upwards positive, rightwards positive) and stick to it throughout the problem. This is crucial for correctly handling vector quantities.
  • ★For graph-based questions, remember that the gradient of a displacement-time graph is velocity, and the gradient of a velocity-time graph is acceleration. The area under a velocity-time graph is displacement.
  • ★For Higher Level projectile motion problems, draw a clear diagram, resolve the initial velocity into its horizontal and vertical components, and treat the horizontal and vertical motions independently.

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