Question

35. A driver of a car moving at 20ms finds a child on the road 10m ahead and brings the
car to the rest just in time to save the child. If the mass of the car with driver is 500 kg,
calculate
i) The force exerted by the brakes on the car.
ii) Time taken to stop the car.​

Answer 1

Answer:

the answer is

i) Force exerted by the brakes of the car is 10000N

ii) time taken to stop the car is 1 second.

Explanation:

initial speed (u) of the car is

$20 {ms}^{ - 1}$

final speed of the car (v) is 0 as it will come to rest after applying the brakes.

distance between the child and car (s) is 10m which is stopping distance of the car.

weight of the car and driver together (m) is 500kg

a is the acceleration.

therefore, applying the formula,

${v}^{2} = {u}^{2} + 2as \\ 0 = ( {20) }^{2} + 2a \times 10 \\ 0 = 400 + 20a \\ 0 - 400 = 20a \\ - \frac{ 400}{20} = a \\ - 20 = a \\ a = - 20 {ms}^{ - 2}$

acceleration here is coming out as negative because the car is performing retardation i.e. breaking.

Now to calculate force,

i) Take only magnitude of acceleration i.e positive value.

$F = ma \\ = 500 \times 20 = 10000 N$

ii)

to calculate the time taken to stop using the formula,

$v = u + at \\ 0 = 20 - 20 \times t \\ 20t = 20 \\ t = \frac{20}{20} \\ t = 1s$

hope it helps