Kinematics – Year 12 A level maths test answers to download as PDF

← Back to the test

1. Answer to Question 1 (4 marks)

   1. List the quantities: \(u=8\), \(v=20\), \(s=140\). The time is unknown, so use \(v^2=u^2+2as\). \(400=64+2a\times140\) so \(a=\dfrac{336}{280}=1.2\) m/s\(^2\) (2) The car accelerates at 1.2 m/s\(^2\).
   2. Use \(v=u+at\) rearranged for \(t\). \(t=\dfrac{20-8}{1.2}=10\) s (2) Check with \(s=\tfrac12(u+v)t=\tfrac12\times28\times10=140\) ✓.

   Mark scheme for (a): 1 for choosing \(v^2=u^2+2as\) with correct values, 1 for 1.2. In (b), \(s=\tfrac12(u+v)t\) gives \(140=14t\), so \(t=10\) s directly, a useful alternative that avoids relying on the acceleration found in (a).

2. Answer to Question 2 (4 marks)

   1. Take upwards as positive; at the top \(v=0\), so use \(v^2=u^2+2as\) with \(a=-9.8\), then add the 2 m starting height. \(h=\dfrac{441}{19.6}=22.5\) m above the start, so 24.5 m above the ground (2) The stone reaches 24.5 m above the ground.
   2. The ground is 2 m below the starting point, so the displacement is \(-2\) m; use \(s=ut+\tfrac12at^2\) and keep the positive root. \(-2=21t-4.9t^2\), \(4.9t^2-21t-2=0\), \(t=\dfrac{21+\sqrt{441+39.2}}{9.8}=4.38\) s (2) The stone hits the ground after about 4.38 s.

3. Answer to Question 3 (4 marks)

   1. The distance is the area under the velocity-time graph: a triangle, a rectangle and another triangle. \(\tfrac12\times60\times30+30\times120+\tfrac12\times40\times30=900+3600+600=5100\) m (3) The train travels 5 100 m, which is 5.1 km.
   2. Divide the total distance by the total time \(60+120+40=220\) s. \(\dfrac{5100}{220}=23.2\) m/s (1) The average speed is about 23.2 m/s.

   Sketch the graph: a rising line from \((0,0)\) to \((60,30)\), a horizontal line to \((180,30)\), then a falling line to \((220,0)\). The trapezium formula gives \(\tfrac12(120+220)\times30=5100\) ✓.

4. Answer to Question 4 (4 marks)

   1. The particle is at rest when \(v=0\); take out the factor 3 and factorise. \(3(t-1)(t-3)=0\), \(t=1\) and \(3\) (2) So it is at rest after 1 s and after 3 s.
   2. Integrate the velocity and use \(s=4\) at \(t=0\) to find the constant. \(s=t^3-6t^2+9t+4\) (1), \(s(4)=64-96+36+4=8\) m (1) At \(t=4\) the particle is 8 m from the origin.

   Between \(t=1\) and \(t=3\) the velocity is negative, so the particle moves backwards.

5. Answer to Question 5 (4 marks)

   1. The police car's distance is \(\tfrac12\times2t^2=t^2\) and the van's distance is \(20t\); they meet when these are equal. \(t^2=20t\) so \(t=20\) (2) The root \(t=0\) is the start, so the police car catches the van after 20 s.
   2. Use \(v=u+at\) with \(u=0\). \(v=2\times20=40\) m/s (1) The police car is moving at 40 m/s, twice the van's speed.
   3. Use the van's distance \(20t\), or the police car's \(t^2\). \(20\times20=400\) m (1) Both give 400 m from the junction.

Total : 20 marks