Year 12 kinematics worksheet answers to download as PDF

These Year 12 kinematics worksheet answers give a full worked solution to all 20 questions of the kinematics worksheet. Try each question yourself first, then check your method step by step.

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Answer to Question 1
1. Acceleration is the gradient of the velocity-time graph: change in velocity divided by time. Stage 1: \(a=\dfrac{12-0}6=\) 2 m/s\(^2\). Stage 3: \(a=\dfrac{0-12}4=-3\), a deceleration of 3 m/s\(^2\).
2. Distance is the area under the velocity-time graph. Stage 1: area of triangle \(\tfrac12\times6\times12=\) 36 m. Stage 2: \(12\times20=\) 240 m. Stage 3: \(\tfrac12\times4\times12=\) 24 m.
3. Total distance \(=36+240+24=\) 300 m in \(6+20+4=30\) s. Average speed \(=\dfrac{300}{30}=\) 10 m/s. The average speed is less than the top speed because the lift spends time speeding up and slowing down.
Answer to Question 2
1. Here \(u\), \(a\) and \(t\) are known, so use \(v=u+at\) and \(s=ut+\tfrac12at^2\). \(v=u+at=3+10=\) 13 m/s. \(s=ut+\tfrac12at^2=15+25=\) 40 m. After 5 s the particle moves at 13 m/s and has travelled 40 m.
2. \(u\), \(v\) and \(s\) are known; \(s=\tfrac12(u+v)t\) gives \(t\) directly. \(s=\tfrac12(u+v)t\): \(66=11t\), so \(t=\) 6 s. \(a=\dfrac{v-u}t=\dfrac{12}6=\) 2 m/s\(^2\). Check with \(v^2=u^2+2as\): \(25+2\times2\times66=289=17^2\) ✓.
3. \(s\), \(t\) and \(a\) are known and \(u\) is wanted, so use \(s=ut+\tfrac12at^2\). \(s=ut+\tfrac12at^2\): \(48=4u+\tfrac12\times1.5\times16=4u+12\), so \(u=\) 9 m/s. Check: \(9\times4+0.75\times16=48\) ✓.
Answer to Question 3
1. List the quantities: \(u=20\), \(v=0\), \(a=-4\); the time is not needed, so use \(v^2=u^2+2as\). \(v^2=u^2+2as\): \(0=400-8s\), so \(s=\) 50 m. The braking distance is 50 m.
2. Use \(v=u+at\) with \(v=0\). \(v=u+at\): \(0=20-4t\), \(t=\) 5 s. The car stops after 5 s.
3. Use \(v^2=u^2+2as\) with \(s=30\). \(v^2=400-2\times4\times30=160\), so \(v=\sqrt{160}=\) 12.6 m/s. After 30 m of braking the car is still moving at about 12.6 m/s.
Answer to Question 4
1. Take upwards as positive, so \(a=-9.8\); at the highest point \(v=0\). Use \(v^2=u^2+2as\). At the top \(v=0\): \(0=196-19.6s\), so \(s=\) 10 m. The ball rises 10 m.
2. Use \(v=u+at\) with \(v=0\); by symmetry the time down equals the time up. \(0=14-9.8t\) gives \(t=1.43\) s up. Total flight \(=2\times1.43=\) 2.86 s. The ball is in the air for about 2.86 s.
3. Substitute \(t=2\) into \(v=u+at\) and \(s=ut+\tfrac12at^2\). \(v=14-9.8\times2=\) \(-5.6\) m/s (5.6 m/s downwards). \(s=14\times2-4.9\times4=28-19.6=\) 8.4 m. The negative velocity shows the ball is already falling, at a height of 8.4 m.
Answer to Question 5
1. Velocity is the derivative of displacement and acceleration is the derivative of velocity. \(v=\dfrac{ds}{dt}=6t^2-18t+12\) and \(a=\dfrac{dv}{dt}=12t-18\).
2. The particle is at rest when \(v=0\); factorise the velocity. \(6(t^2-3t+2)=6(t-1)(t-2)=0\) at \(t=1\) and \(t=2\). So the particle is at rest at \(t=1\) and \(t=2\).
3. Substitute the times into \(s\) and \(a\). \(s(1)=2-9+12=\) 5 m, \(s(2)=16-36+24=\) 4 m. \(a(1)=12-18=\) \(-6\) m/s\(^2\). At \(t=1\) the particle is 5 m from the origin and its velocity is decreasing.
Answer to Question 6
1. Velocity is the integral of acceleration; the initial velocity gives the constant. \(v=\int(6t-4)dt=3t^2-4t+c\). \(v(0)=3\) so \(c=3\): \(v=3t^2-4t+3\). This is the velocity at time \(t\).
2. Displacement is the integral of velocity; the initial displacement gives the constant. \(s=\int v\,dt=t^3-2t^2+3t+d\). \(s(0)=0\) so \(d=0\): \(s=t^3-2t^2+3t\). This is the displacement at time \(t\).
3. Substitute \(t=2\) into both expressions. \(v(2)=12-8+3=\) 7 m/s. \(s(2)=8-8+6=\) 6 m. At \(t=2\) the particle moves at 7 m/s and is 6 m from its starting point.
Answer to Question 7
1. Use \(t=\dfrac{v-u}a\) for the acceleration and deceleration stages; 4 minutes is 240 s. Stage 1: \(t=\dfrac{25}{0.5}=50\) s. Stage 2: \(4\times60=240\) s. Stage 3: \(t=\dfrac{25}1=25\) s. Total 315 s.
2. The distance is the area under the velocity-time graph, made of two triangles and a rectangle. Stage 1: \(\tfrac12\times25\times50=625\) m. Stage 2: \(25\times240=6000\) m. Stage 3: \(\tfrac12\times25\times25=312.5\) m. Total 6 937.5 m. The stations are about 6.94 km apart.
3. Average speed is total distance divided by total time; multiply by 3.6 to convert m/s to km/h. Average speed \(=\dfrac{6937.5}{315}=\) 22.0 m/s \(=22.02\times3.6=\) 79.3 km/h.
4. A trapezium: rising line from \((0,0)\) to \((50,25)\), horizontal to \((290,25)\), falling to \((315,0)\). The area of this trapezium is the distance found in part (b).
Answer to Question 8
1. During the reaction time the car moves at constant speed, so distance \(=\) speed \(\times\) time. \(15\times0.7=\) 10.5 m. The thinking distance is 10.5 m.
2. Use \(v^2=u^2+2as\) with \(u=15\), \(v=0\), \(a=-5\). \(0=225-10s\), so \(s=\) 22.5 m. The braking distance is 22.5 m.
3. Compare the total stopping distance with the distance to the hazard. Total stopping distance \(=10.5+22.5=33\) m \(>30\) m, so she hits the hazard.
4. Only the distance left after the thinking distance is available for braking; use \(v^2=u^2+2as\). Braking distance available \(=30-10.5=19.5\) m. \(v^2=225-2\times5\times19.5=30\), so \(v=\) 5.48 m/s. She hits the hazard at about 5.48 m/s, much slower than her original speed.
Answer to Question 9
1. Take downwards as positive with \(u=0\) and \(a=9.8\); use \(s=ut+\tfrac12at^2\), then \(v=u+at\). \(45=\tfrac12\times9.8t^2\), \(t^2=9.18\), \(t=\) 3.03 s. \(v=9.8\times3.03=\) 29.7 m/s. The stone takes about 3.03 s and hits the water at about 29.7 m/s.
2. Now \(u=5\) (downwards) with \(a=9.8\) and \(s=45\). \(s=ut+\tfrac12at^2\): \(45=5t+4.9t^2\), so \(4.9t^2+5t-45=0\). This is the required equation.
3. Use the quadratic formula and take the positive root, then use \(v=u+at\). \(t=\dfrac{-5+\sqrt{25+882}}{9.8}=\dfrac{-5+30.12}{9.8}=\) 2.56 s. \(v=5+9.8\times2.563=\) 30.1 m/s. The second stone arrives about 0.47 s sooner and slightly faster.
Answer to Question 10
1. Use \(v=u+at\) with \(u=0\). \(v=1.2\times3=\) 3.6 m/s. The maximum speed is 3.6 m/s.
2. The lift slows from 3.6 m/s to 0 in 4 s. \(a=\dfrac{0-3.6}4=-0.9\): deceleration 0.9 m/s\(^2\).
3. The height climbed is the area under the velocity-time graph (two triangles and a rectangle). Distances: \(\tfrac12\times3\times3.6=5.4\) m; \(3.6\times10=36\) m; \(\tfrac12\times4\times3.6=7.2\) m. Total 48.6 m. The lift climbs 48.6 m.
4. Divide the total distance by the total time. Time \(=3+10+4=17\) s. Average speed \(=\dfrac{48.6}{17}=\) 2.86 m/s. The average speed is about 2.86 m/s.
Answer to Question 11
1. \(A\) starts 12 m ahead of \(B\)'s starting point and moves at a constant 7 m/s. \(12+7t\). This is \(A\)'s distance from \(B\)'s starting point.
2. \(B\) starts from rest, so use \(s=\tfrac12at^2\). \(s=\tfrac12\times1.5t^2=0.75t^2\). This is \(B\)'s distance from her starting point.
3. \(B\) catches \(A\) when both distances are equal; solve with the formula and keep the positive root. \(0.75t^2=12+7t\), so \(0.75t^2-7t-12=0\). \(t=\dfrac{7+\sqrt{49+36}}{1.5}=\dfrac{7+9.22}{1.5}=\) 10.8 s. \(B\) catches \(A\) after about 10.8 s.
4. Substitute \(t=10.81\) into \(s=0.75t^2\) and \(v=at\). \(s=0.75\times10.81^2=\) 87.7 m. \(v=1.5\times10.81=\) 16.2 m/s, faster than \(A\). \(B\) has run about 87.7 m when she catches \(A\).
Answer to Question 12
1. Integrate the acceleration and use the initial speed to find the constant. \(v=\int(0.6-0.04t)dt=0.6t-0.02t^2+c\), \(v(0)=2\). This is the required expression.
2. Substitute \(t=15\) into \(v\) and into \(a\). \(v(15)=2+9-4.5=\) 6.5 m/s. \(a(15)=0.6-0.6=0\): the lorry has reached its maximum speed.
3. The distance is the integral of the velocity from 0 to 15 (the velocity is always positive here). \(s=\int_0^{15}(2+0.6t-0.02t^2)dt=\left[2t+0.3t^2-\dfrac{0.02}3t^3\right]_0^{15}=30+67.5-22.5=\) 75 m. The lorry travels 75 m.
4. Divide the total distance by the total time. \(\dfrac{75}{15}=\) 5 m/s. The average speed is 5 m/s.
Answer to Question 13
1. At the highest point \(v=0\); use \(v^2=u^2+2as\) and \(v=u+at\) with \(a=-9.8\). \(0=1600-19.6s\), \(s=\) 81.6 m. \(t=\dfrac{40}{9.8}=\) 4.08 s. The rocket explodes 81.6 m up, about 4.08 s after launch.
2. Substitute \(t=3\) into \(s=ut+\tfrac12at^2\). \(s=40\times3-4.9\times9=120-44.1=\) 75.9 m. After 3 s the rocket is 75.9 m above the ground.
3. \(40t-4.9t^2>70\Rightarrow4.9t^2-40t+70<0\). \(t=\dfrac{40\pm\sqrt{1600-1372}}{9.8}=\dfrac{40\pm15.1}{9.8}\), giving \(t=2.54\) and \(5.62\). Time above 70 m \(=5.62-2.54=\) 3.08 s. (The ascent and descent each pass 70 m: the rocket is above 70 m on both sides of the top.)
Answer to Question 14
1. Use \(s=ut+\tfrac12at^2\) with \(u=20\), \(a=0.8\), \(s=300\). \(300=20t+\tfrac12\times0.8t^2=20t+0.4t^2\). Divide by 0.4: \(t^2+50t-750=0\). This is the required equation.
2. Use the quadratic formula and take the positive root. \(t=\dfrac{-50+\sqrt{2500+3000}}2=\) 12.08 s. The car takes about 12.1 s.
3. Use \(v=u+at\). \(v=20+0.8\times12.08=\) 29.7 m/s. The final speed is about 29.7 m/s.
4. This equation does not use \(t\), so it is an independent check. \(v^2=400+2\times0.8\times300=880\), \(v=\sqrt{880}=29.66\) ✓. Both methods give the same final speed.
Answer to Question 15
1. Acceleration is the change in velocity divided by the time taken. Stage 1: \(\dfrac68=0.75\) m/s\(^2\). Stage 3: \(\dfrac{6-2}{10}=0.4\) m/s\(^2\) deceleration.
2. The distance in each stage is the area under that part of the velocity-time graph. Stage 1: \(\tfrac12\times8\times6=24\) m. Stage 2: \(6\times40=240\) m. Stage 3: \(\tfrac{6+2}2\times10=40\) m. Stage 4: \(2\times30=60\) m.
3. Add the distances and divide by the total time \(8+40+10+30=88\) s. Total \(=24+240+40+60=364\) m in 88 s. Average speed \(=\dfrac{364}{88}=\) 4.14 m/s.
4. 4.14 m/s is a little more than 4 m/s, so she is right to the nearest m/s but slightly underestimates.
Answer to Question 16
1. Integrate \(v\) and use \(s=0\) at \(t=0\). \(s=\int(t^2-8t+12)dt=\dfrac{t^3}3-4t^2+12t\) (since \(s(0)=0\)). This is the required expression.
2. Substitute each time into \(s\). \(s(2)=\dfrac83-16+24=\dfrac{32}3=10.67\) m. \(s(6)=72-144+72=0\). \(s(8)=\dfrac{512}3-256+96=\dfrac{32}3=10.67\) m.
3. The particle changes direction at \(t=2\) and \(t=6\) (where \(v=0\)), so add the distances of each stage. Forward 10.67 m, then back 10.67 m, then forward 10.67 m: total 32 m. The total distance is 32 m, although the final displacement is only 10.67 m.
4. Acceleration is the derivative of velocity. \(a=2t-8=0\) at \(t=4\). The velocity has its minimum (\(-4\) m/s) there, so the acceleration changes sign: negative before and positive after.
Answer to Question 17
1. Use \(s=ut+\tfrac12at^2\) for the motion from \(A\) to \(B\) and from \(A\) to \(C\). \(60=4u+8a\), i.e. \(u+2a=15\). \(210=10u+50a\), i.e. \(u+5a=21\).
2. Subtract the equations to eliminate \(u\). Subtract: \(3a=6\), so \(a=2\) and \(u=11\). So the car passes \(A\) at 11 m/s with acceleration 2 m/s\(^2\).
3. Use \(v=u+at\) with \(t=4\) and \(t=10\). At \(B\): \(11+8=\) 19 m/s. At \(C\): \(11+20=\) 31 m/s.
4. The time from \(B\) to \(C\) is \(10-4=6\) s. \(210-60=\) 150 m. Time 6 s. Average speed \(=25\) m/s (the mean of 19 and 31). For constant acceleration the average speed is the mean of the initial and final speeds.
Answer to Question 18
1. Use \(v^2=u^2+2as\) with \(u=12\), \(v=0\), \(a=-9.8\). \(0=144-19.6s\), \(s=\) 7.35 m. The ball rises 7.35 m above the cliff top.
2. The sea is 25 m below the starting point, so the displacement is \(-25\) m. Displacement \(-25=12t-4.9t^2\), so \(4.9t^2-12t-25=0\). \(t=\dfrac{12+\sqrt{144+490}}{9.8}=\dfrac{12+25.18}{9.8}=\) 3.79 s.
3. Use \(v=u+at\) with the time from part (b). \(v=12-9.8\times3.794=-25.2\): speed 25.2 m/s. The ball hits the sea at about 25.2 m/s.
4. The ball goes up, comes back down to the cliff top and then falls 25 m to the sea. Up 7.35 m, down 7.35 m (back to the cliff), then down 25 m: total \(=14.7+25=\) 39.7 m. The total distance is 39.7 m, while the displacement is only 25 m downwards.
Answer to Question 19
1. Use \(v=u+at\) with \(u=V\), \(v=0\), \(a=-0.5\). \(0=V-0.5t\), so \(t=2V\). This is the required time.
2. Find the area under each part of the velocity-time graph. Stage 1: \(\tfrac12\times40\times V=20V\). Stage 3: \(\tfrac12\times2V\times V=V^2\). Stage 2 lasts \(300-40-2V=260-2V\) s, covering \(V(260-2V)=260V-2V^2\). Total \(=20V+260V-2V^2+V^2=280V-V^2\). This is the required expression.
3. Set the total distance equal to 8000 m and solve the quadratic. \(280V-V^2=8000\Rightarrow V^2-280V+8000=0\Rightarrow V=\dfrac{280-\sqrt{46400}}2=\) 32.3 m/s (the other root, 247.7, would make stage 2 negative).
4. Use \(a=\dfrac V{40}\) and the time \(260-2V\) for the constant speed stage. \(a=\dfrac{32.3}{40}=\) 0.81 m/s\(^2\). Constant speed time \(=260-64.6=\) 195.4 s.
Answer to Question 20
1. Take upwards as positive for both particles; both have acceleration \(-9.8\). Height of \(A\): \(80-4.9t^2\). Height of \(B\): \(ut-4.9t^2\). They meet when these are equal: \(80=ut\), so \(t=\dfrac{80}u\). (The \(4.9t^2\) terms cancel because both objects accelerate equally.) So the meeting time does not depend on \(g\).
2. Substitute \(u=40\) into \(t=\dfrac{80}u\), then find the height of either particle. \(t=\dfrac{80}{40}=\) 2 s. Height \(=40\times2-4.9\times4=80-19.6=\) 60.4 m. They meet 2 s after release, 60.4 m above the ground.
3. Set the height of \(A\) at the meeting time equal to 20 m. Height \(=80-4.9\left(\dfrac{80}u\right)^2=20\), so \(4.9\times\dfrac{6400}{u^2}=60\), \(u^2=\dfrac{31360}{60}=522.7\), so \(u=\) 22.9 m/s. Check: \(t=\dfrac{80}{22.9}=3.5\) s and \(80-4.9\times3.5^2=20\) ✓.

How to use these Year 12 kinematics worksheet answers

First, try every question on the worksheet without looking. Then compare your working with the solution line by line, not only the final answer. When your answer is different, find the first step that changes, because that is where the mistake is. After that, read the matching part of the kinematics lesson again. Finally, check that the method is secure with the quiz or the test.

Also, the Year 12 kinematics worksheet answers are available as a PDF, so you can mark printed work quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, these Year 12 kinematics worksheet answers show every step, so nothing is left unexplained.