What is covered in Grade 10 Physical Sciences Equations of Motion?

The four kinematic equations relate displacement, initial velocity, final velocity, acceleration, and time for uniformly accelerated motion. They can only be used when acceleration is constant.

What is Instantaneous speed in Physical Sciences?

The speed of an object at a specific instant in time.

What is Instantaneous velocity?

The velocity of an object at a specific instant in time. It is the gradient of the tangent to the position-time graph at that point.

What is the formula for Equation 1 (v-t relationship)?

The formula for Equation 1 (v-t relationship) is: vf = vi + aΔt. Where: vf = final velocity (m·s⁻¹), vi = initial velocity (m·s⁻¹), a = acceleration (m·s⁻²), Δt = time (s) The SI unit is m·s⁻¹.

Is Grade 10 Physical Sciences Equations of Motion on the CAPS curriculum?

Yes. Equations of Motion is part of the official CAPS Physical Sciences curriculum for Grade 10, Term 3 in South Africa. The content on MathSciBuddy follows the Annual Teaching Plan (ATP) set by the Department of Basic Education.

Equations of Motion Grade 10 Physical Sciences CAPS Notes

When acceleration is constant, four powerful equations relate five variables: displacement (Δx), initial velocity (vᵢ), final velocity (vf), acceleration (a), and time (Δt). Master the 'known–unknown' approach: identify what you have, identify what you want, pick the equation that links them.

Week 7–8

The Four Equations of Motion

State and apply all four equations of motionSolve problems for objects under uniform accelerationSolve free-fall problems using g = 9,8 m·s⁻²

Definition

Uniformly accelerated motion

Uniformly accelerated motion is motion in which the acceleration is constant (does not change with time).

Definition

Free fall

Free fall is the motion of an object under the influence of gravity only, with no air resistance; acceleration = g = 9,8 m·s⁻² directed downward.

Formula

Equation 1 — velocity and time

v_f = v_i + a\Delta t

vf = final velocity (m·s⁻¹), vᵢ = initial velocity (m·s⁻¹), a = acceleration (m·s⁻²), Δt = time (s)

SI unit: m·s⁻¹

Formula

Equation 2 — displacement using average velocity

\Delta x = \frac{v_i + v_f}{2} \cdot \Delta t

Δx = displacement (m), vᵢ and vf in m·s⁻¹, Δt = time (s)

SI unit: m

Formula

Equation 3 — displacement from initial velocity

\Delta x = v_i \Delta t + \frac{1}{2}a\Delta t^2

Δx = displacement (m), vᵢ = initial velocity (m·s⁻¹), a = acceleration (m·s⁻²), Δt = time (s)

SI unit: m

Formula

Equation 4 — velocity and displacement (no time)

v_f^2 = v_i^2 + 2a\Delta x

vf = final velocity (m·s⁻¹), vᵢ = initial velocity (m·s⁻¹), a = acceleration (m·s⁻²), Δx = displacement (m)

SI unit: (m·s⁻¹)²

✓

Exam Tip

Choose the equation that contains the variable you want to find AND all three known variables. The 'missing variable' technique: which variable appears in none of what you know? Use the equation that does NOT contain it.

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Watch Out

ALL four equations only apply when acceleration is CONSTANT. Do not use them if acceleration is changing. Always define a positive direction before solving — be consistent throughout the problem.

Worked Example

A car starts from rest and accelerates at 3 m·s⁻² for 8 s. Calculate: (a) final velocity, (b) distance covered.

Given

vᵢ = 0 m·s⁻¹ (starts from rest)
a = 3 m·s⁻²
Δt = 8 s

Find

(a) vf (b) Δx

Solution

1(a) Use Equation 1: vf = vᵢ + aΔt
2vf = 0 + 3 × 8 = 24 m·s⁻¹
3(b) Use Equation 3: Δx = vᵢΔt + ½aΔt²
4Δx = 0(8) + ½(3)(8²) = 0 + ½(3)(64) = 96 m

Answer: (a) vf = 24 m·s⁻¹; (b) Δx = 96 m

Worked Example

A car decelerates from 30 m·s⁻¹ to rest over 60 m. Calculate: (a) acceleration, (b) time to stop.

Given

vᵢ = 30 m·s⁻¹
vf = 0 m·s⁻¹
Δx = 60 m

Find

(a) a (b) Δt

Solution

1(a) Use Equation 4: vf² = vᵢ² + 2aΔx
20 = 30² + 2a(60)
30 = 900 + 120a
4a = −900/120 = −7,5 m·s⁻²
5(b) Use Equation 1: vf = vᵢ + aΔt
60 = 30 + (−7,5)Δt
7Δt = 30/7,5 = 4 s

Answer: (a) a = −7,5 m·s⁻² (deceleration); (b) Δt = 4 s

Worked Example

A ball is thrown upward at 20 m·s⁻¹ from the ground. (g = 9,8 m·s⁻²) (a) Time to reach maximum height. (b) Maximum height.

Given

vᵢ = +20 m·s⁻¹ (upward, defined positive)
a = −9,8 m·s⁻² (gravity acts downward)
vf = 0 m·s⁻¹ at maximum height

Find

(a) Δt to max height (b) Δx at max height

Solution

1(a) Use Equation 1: vf = vᵢ + aΔt
20 = 20 + (−9,8)Δt
3Δt = 20/9,8 = 2,04 s
4(b) Use Equation 4: vf² = vᵢ² + 2aΔx
50 = 400 + 2(−9,8)Δx
619,6Δx = 400
7Δx = 400/19,6 = 20,4 m

Answer: (a) t = 2,04 s; (b) h = 20,4 m

Practice Question

A train accelerates from 5 m·s⁻¹ to 25 m·s⁻¹ in 10 s. Calculate: (a) acceleration, (b) distance travelled.

(6 marks)

Practice Question

A stone is dropped from a bridge 19,6 m above the water. How long does it take to reach the water? (g = 9,8 m·s⁻², vᵢ = 0)

(4 marks)

Practice Question

A car travelling at 25 m·s⁻¹ applies brakes and decelerates at 5 m·s⁻² to rest. Calculate the stopping distance.

(4 marks)