A car is travelling at 20 m/s. The driver sees a hazard and has a reaction time of 0.70 s. The brakes then give a constant deceleration of 5.0 m/s2.
Use the equation:
(final velocity)2−(initial velocity)2=2×acceleration×distance(\text{final velocity})^2 - (\text{initial velocity})^2 = 2 \times \text{acceleration} \times \text{distance}(final velocity)2−(initial velocity)2=2×acceleration×distance
What is the stopping distance of the car?
14 m14\ \text{m}14 m
54 m54\ \text{m}54 m
40 m40\ \text{m}40 m
94 m94\ \text{m}94 m
50 exam-style questions on OCR GCSE Physics 8.1 Physics on the move, covering 8.1.1 Typical everyday speeds, 8.1.2 Typical everyday accelerations, 8.1.3 Ratios, units and rates on the road, 8.1.4 Human reaction times, 8.1.5 Factors affecting stopping distance, 8.1.6 Stopping distances at different speeds, 8.1.7 Dangers of large decelerations, 8.1.8 Estimating forces in road situations, and 8.1.9 Estimating the effects of large accelerations. Each one has a worked solution and a mark scheme showing where the marks go.