A car driver makes an emergency stop.
The thinking distance is 12 m12\text{ m}12 m and the braking distance is 26 m26\text{ m}26 m.
Calculate the total stopping distance of the car.
In another scenario, a car travels at a speed of 24 m/s24\text{ m/s}24 m/s. After the brakes are fully applied, it takes 6.0 s6.0\text{ s}6.0 s for the car to come to a complete stop.
Calculate the deceleration of the car. Use the equation:
acceleration=change in velocitytime \text{acceleration} = \frac{\text{change in velocity}}{\text{time}} acceleration=timechange in velocityThe braking system of this car is upgraded to a high-performance system. The same car travelling at 24 m/s24\text{ m/s}24 m/s now takes only 0.60 s0.60\text{ s}0.60 s to stop after the brakes are applied.
A passenger claims: "Because the stopping time is ten times shorter, this upgraded braking system makes the passenger ten times safer in an emergency stop."
Explain why this claim is incorrect, referencing the forces acting on the passenger.
Suggest one safety feature in a car designed to reduce passenger injury caused by these forces during rapid deceleration.
27 exam-style questions on OCR GCSE Physics Physics on the move. Each one has a worked solution and a mark scheme showing where the marks go.