The problems are on the suvat equations worksheet. Try them on your own first, then compare with the answers below.
Quick answer: The suvat equations connect displacement, initial velocity, final velocity, acceleration and time for motion in a straight line with constant acceleration. These worked answers correct the 20 problems of the Physics 11-12 worksheet on suvat equations, from choose the equation to chasing stones: each solution shows the formula, the substitution and the final answer with its unit.
Each answer lists the equation, the substitution and the result in bold. Use g = 9.81 m/s². Small differences in the last digit are acceptable when your rounding differs. Compare your choice of equation as well as your numbers.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.4, IB A.1
Problem 1 Choose the equation
1. s = ut + ½at².2. v² = u² + 2as.
3. s = ½(u + v)t.
4. v = u + at.
Problem 2 Tram departure
1. v = u + at = 0 + 1.2 × 15 = 18 m/s.2. 18 × 3.6 = 64.8 km/h.
Problem 3 Cyclist speeding up
1. a = (10.0 − 4.0) ÷ 12 = 0.50 m/s².2. s = ½(4.0 + 10.0) × 12 = 84 m.
Problem 4 Falling mango
1. t = √(2s ÷ g) = √(10.0 ÷ 9.81) = 1.01 s.2. v = gt = 9.81 × 1.01 = 9.9 m/s.
Problem 5 Emergency stop
1. 90 ÷ 3.6 = 25 m/s.2. a = (0 − 25) ÷ 6.0 = −4.2 m/s².
3. s = ½ × 25 × 6.0 = 75 m.
Problem 6 Train journey graph
1. a = 20 ÷ 10 = 2.0 m/s².2. Area = ½ × 10 × 20 + 20 × 20 + ½ × 5 × 20 = 100 + 400 + 50 = 550 m.
3. a = (0 − 20) ÷ 5 = −4.0 m/s².
Problem 7 Stopping distance
1. s = 20 × 0.70 = 14.0 m.2. v² = u² + 2as: 0 = 400 − 2 × 6.5 × s, so s = 400 ÷ 13 = 30.8 m.
3. 14.0 + 30.8 = 44.8 m.
Problem 8 Sprint start
1. s = ½at², so a = 2s ÷ t² = 20.0 ÷ 4.84 = 4.13 m/s².2. v = at = 4.13 × 2.20 = 9.09 m/s.
Problem 9 Well depth
1. s = ½gt² = 0.5 × 9.81 × 2.4² = 28.3 m.2. v = gt = 9.81 × 2.4 = 23.5 m/s.
Problem 10 Ball thrown down
1. v² = u² + 2as = 25 + 2 × 9.81 × 12 = 260.4, so v = 16.1 m/s.2. t = (v − u) ÷ g = (16.14 − 5.0) ÷ 9.81 = 1.14 s.
Problem 11 Tennis ball thrown up
1. h = u² ÷ 2g = 324 ÷ 19.62 = 16.5 m.2. t = u ÷ g = 18 ÷ 9.81 = 1.83 s.
3. Total = 2 × 1.835 = 3.67 s.
Problem 12 Ruler reaction test
1. 18.0 cm = 0.180 m.2. t = √(2s ÷ g) = √(0.360 ÷ 9.81) = √0.03670 = 0.192 s.
Problem 13 Lift motion
1. a = 3.0 ÷ 2.5 = 1.2 m/s².2. Area = ½ × 2.5 × 3.0 + 6.0 × 3.0 + ½ × 2.5 × 3.0 = 3.75 + 18.0 + 3.75 = 25.5 m.
3. Time = 11 s, so average speed = 25.5 ÷ 11 = 2.3 m/s.
Problem 14 Spot the sign error
1. She used a positive acceleration while taking up as positive. The acceleration is downwards, so a = −9.81 m/s². The sign of a is wrong.2. v = 10 − 9.81 × 0.50 = 5.1 m/s upwards.
Problem 15 Speeding up over a distance
1. a = (v² − u²) ÷ 2s = (400 − 64) ÷ 300 = 1.12 m/s².2. t = (20 − 8.0) ÷ 1.12 = 10.7 s.
Problem 16 Overtaking
1. A: s = 18t. B: s = ½ × 3.0 × t² = 1.5t².2. 18t = 1.5t², so t = 18 ÷ 1.5 = 12 s.
3. s = 18 × 12 = 216 m; v = 3.0 × 12 = 36 m/s, twice A’s speed.
Problem 17 Stone from a cliff
1. h = 64 ÷ 19.62 = 3.26 m.2. Take up as positive, s = −40 m: v² = 64 + 2 × 9.81 × 40 = 848.8, so v = 29.1 m/s downwards.
3. t = (8.0 + 29.13) ÷ 9.81 = 3.79 s.
Problem 18 Trolley data
1. s ÷ t² = 1.25 for every pair (0.05 ÷ 0.04 = 1.25; 1.25 ÷ 1.00 = 1.25), so s ∝ t².2. From s = ½at², the gradient equals ½a.
3. a = 2 × 1.25 = 2.5 m/s².
4. s = 1.25 × 1.96 = 2.45 m.
Problem 19 Late braking
1. s = u² ÷ 2a = 900 ÷ 16 = 56.3 m.2. Reaction distance = 15 m, total = 71.3 m, which is less than 100 m: it stops about 29 m before the tree.
3. Reaction distance = 45 m, so 55 m remain: v² = 900 − 2 × 8.0 × 55 = 20, v = 4.5 m/s.
Problem 20 Chasing stones
1. First: s = 4.905t². Second: s = 15(t − 1) + 4.905(t − 1)².2. Equating: 4.905t² = 15t − 15 + 4.905t² − 9.81t + 4.905, so 5.19t = 10.095 and t = 1.95 s.
3. s = 4.905 × 1.945² = 18.6 m.
