A car travels \(150\,\mathrm{m}\) in \(12\,\mathrm{s}\). Calculate its average speed.
\(v = s/t = 150/12 = 12.5\,\mathrm{m\,s}^{-1}\).
Forces & Motion
10 exam-style questions on speed, acceleration, Newton’s laws, weight and stopping distances. Tap Show answer for full working.
3 Easy4 Medium3 Hard
Calculate the weight of a \(55\,\mathrm{kg}\) person on Earth. (\(g = 9.8\,\mathrm{N\,kg}^{-1}\).)
\(W = mg = 55\times9.8 = 539\,\mathrm{N}\).
State Newton’s first and second laws of motion.
First: an object remains at rest or at constant velocity unless acted on by a resultant force.
Second: resultant force \(F = ma\) (acceleration is proportional to resultant force and inversely proportional to mass).
A cyclist accelerates from \(2.0\,\mathrm{m\,s}^{-1}\) to \(8.0\,\mathrm{m\,s}^{-1}\) in \(4.0\,\mathrm{s}\). Calculate the acceleration.
\(a = (v-u)/t = (8-2)/4 = 1.5\,\mathrm{m\,s}^{-2}\).
A resultant force of \(24\,\mathrm{N}\) acts on a \(6.0\,\mathrm{kg}\) object. Calculate the acceleration and the distance travelled from rest in \(3.0\,\mathrm{s}\).
\(a = F/m = 24/6 = 4.0\,\mathrm{m\,s}^{-2}\).
\(s = ut + \frac12 at^{2} = 0 + 0.5\times4\times9 = 18\,\mathrm{m}\).
A car of mass \(1000\,\mathrm{kg}\) is travelling at \(20\,\mathrm{m\,s}^{-1}\). The brakes provide a constant decelerating force of \(4000\,\mathrm{N}\). Calculate the braking distance.
\(a = F/m = -4000/1000 = -4.0\,\mathrm{m\,s}^{-2}\).
\(v^{2} = u^{2} + 2as \Rightarrow 0 = 400 + 2(-4)s \Rightarrow s = 50\,\mathrm{m}\).
Explain the difference between thinking distance and braking distance, and state one factor that affects each.
Thinking distance: distance travelled during the driver’s reaction time (affected by speed, tiredness, alcohol, distractions).
Braking distance: distance travelled while the brakes are applied (affected by speed, road conditions, tyre/brake condition, mass of vehicle).
A rocket of mass \(500\,\mathrm{kg}\) produces a thrust of \(8000\,\mathrm{N}\). Calculate the initial acceleration as it lifts off. (\(g = 9.8\,\mathrm{N\,kg}^{-1}\).) Explain why the acceleration increases as fuel is used.
Weight \(= 500\times9.8 = 4900\,\mathrm{N}\). Resultant force upward \(= 8000 - 4900 = 3100\,\mathrm{N}\).
\(a = 3100/500 = 6.2\,\mathrm{m\,s}^{-2}\).
As fuel is used, mass decreases, so for the same thrust the acceleration increases (\(a = F/m\)).
A velocity–time graph shows a vehicle accelerating uniformly from rest to \(15\,\mathrm{m\,s}^{-1}\) in \(5\,\mathrm{s}\), then travelling at constant speed for \(10\,\mathrm{s}\), then decelerating uniformly to rest in \(3\,\mathrm{s}\). Calculate the total distance travelled.
Area under graph: triangle \(= 0.5\times5\times15 = 37.5\,\mathrm{m}\); rectangle \(= 15\times10 = 150\,\mathrm{m}\); triangle \(= 0.5\times3\times15 = 22.5\,\mathrm{m}\).
Total \(= 210\,\mathrm{m}\).
Two forces of \(12\,\mathrm{N}\) and \(5.0\,\mathrm{N}\) act at right angles to each other on an object. Calculate the magnitude of the resultant force.
Resultant \(= \sqrt{12^{2}+5^{2}} = \sqrt{144+25} = \sqrt{169} = 13\,\mathrm{N}\).