Motion in One Dimension (Grade 10 NSC Matric Physical Sciences): Revision Notes

Worked Example: Finding Acceleration

Question: A racing car travels north and accelerates uniformly, covering a distance of 725 m in 10 s. If it has an initial velocity of 10 m·s⁻¹, find its acceleration.

Solution:

Step 1 - Identify given information:

• $v_i = 10$ m·s⁻¹
• $\Delta x = 725$ m
• $t = 10$ s
• $a = ?$ (unknown)

Step 2 - Choose the appropriate equation:

• Since we know $v_i$, $\Delta x$, and $t$, but need to find $a$, we use Equation 3:
• $\Delta x = v_i t + \frac{1}{2}at^2$

Step 3 - Substitute values and solve:

• $725 = (10 \times 10) + \frac{1}{2}a \times (10)^2$
• $725 = 100 + \frac{1}{2}a \times 100$
• $725 - 100 = 50a$
• $625 = 50a$
• $a = 12.5 { m·s⁻²}$

Step 4 - Final answer: The racing car is accelerating at 12.5 m·s⁻² north.

Worked Example: Multiple Motion Calculations

Question: A motorcycle travelling east starts from rest and moves in a straight line with constant acceleration, covering a distance of 64 m in 4 s. Calculate:

• its acceleration
• its final velocity
• at what time the motorcycle had covered half the total distance
• what distance the motorcycle had covered in half the total time

Solution:

Finding acceleration: Given: $v_i = 0$ m·s⁻¹ (starts from rest), $\Delta x = 64$ m, $t = 4$ s

Using Equation 3: $\Delta x = v_i t + \frac{1}{2}at^2$

• $64 = (0 \times 4) + \frac{1}{2}a \times (4)^2$
• $64 = \frac{1}{2}a \times 16$
• $a = 8 { m·s⁻² east}$

Finding final velocity:

• Using Equation 1: $v_f = v_i + at$
• $v_f = 0 + (8)(4) = 32 { m·s⁻¹ east}$

Time at half distance (32 m):

• Using Equation 3: $32 = 0 + \frac{1}{2}(8)t^2$
• $32 = 4t^2$
• $t = 2.83 \text{ s}$

Distance at half time (2 s):

• Using Equation 3: $\Delta x = 0 + \frac{1}{2}(8)(2)^2$
• $\Delta x = 4 \times 4 = 16 \text{ m east}$

Worked Example: Stopping Distance Analysis

Question: A truck travels at a constant velocity of 10 m·s⁻¹ when the driver sees a child 50 m ahead. The driver's reaction time is 0.5 s, after which the truck brakes with an acceleration of -1.25 m·s⁻². Will the truck hit the child?

Solution:

This problem has two phases: constant velocity during reaction time, then deceleration during braking.

Phase 1 - Constant velocity (reaction time): Distance = velocity × time = $10 \times 0.5 = 5$ m

Phase 2 - Braking (deceleration): Given: $v_i = 10$ m·s⁻¹, $v_f = 0$ m·s⁻¹, $a = -1.25$ m·s⁻²

First, find braking time using Equation 1: $0 = 10 + (-1.25)t$ $t = 8 \text{ s}$

Then find braking distance using Equation 2: $\Delta x = \frac{(v_i + v_f)t}{2} = \frac{(10 + 0) \times 8}{2} = 40 \text{ m}$

Total stopping distance: $5 + 40 = 45$ m

Since the child is 50 m ahead and the truck stops in 45 m, the truck will not hit the child.