9702 · 2.1

Equations of motion — practice questions

Practice and worked examples for 9702 Equations of motion. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A car accelerates uniformly from rest to a speed of $20.0 \, \text{m/s}$ in $5.00 \, \text{s}$ . Calculate the distance it travels during this time.

Show solution outline

Identify knowns: Initial velocity ( $u$ ) = $0 \, \text{m/s}$ (from rest) Final velocity ( $v$ ) = $20.0 \, \text{m/s}$ Time ( $t$ ) = $5.00 \, \text{s}$
Identify unknown: Displacement ( $s$ ).
Choose appropriate SUVAT equation: The equation $s = \frac{1}{2}(u+v)t$ includes $u, v, t$ and allows us to find $s$ .
Substitute values: $s = \frac{1}{2}(0 + 20.0)(5.00)$
Calculate: $s = \frac{1}{2}(20.0)(5.00) = 10.0 \times 5.00 = 50.0 \, \text{m}$ The car travels $50.0 \, \text{m}$ .

Worked example 2

A ball is thrown vertically upwards from the ground with an initial speed of 15.0 m/s. Ignoring air resistance, calculate the maximum height it reaches. (Take g = 9.81 m/s²)

Show solution outline

Define direction and list knowns: Let's define the upward direction as positive. Initial velocity (u) = +15.0 m/s At maximum height, the final velocity (v) = 0 m/s Acceleration (a) = -g = -9.81 m/s² (it's downwards)
Identify unknown: Maximum height, which is the displacement (s).
Choose appropriate SUVAT equation: We have u, v, a and need to find s. The equation that links these without time (t) is v² = u² + 2as.
Rearrange and substitute: Rearrange the formula to solve for s: s = (v² - u²) / 2a s = (0² - (15.0)²) / (2 × -9.81)
Calculate: s = (-225) / (-19.62) s = 11.467... m
Final Answer: The maximum height reached by the ball is 11.5 m (to 3 significant figures).

Mark a 9702 question on this topic Full lesson Open in course studio