Question

A car is traveling 20 m s when the driver sees a child standing in the road. he takes 0.8 seconds to react, then steps on the brakes and slows down at 7.0 m/s squared. how far does the car go before it stops?

Answer 1

We use kinematic equations,

During reaction time,

$S = ut + \frac{1}{2} at^2$                        (A)

During the deceleration,

$v^2= u^2 + 2as$.                                   (B)

Here, u is initial velocity, v is final velocity , a is acceleration and t is time.

There is no deceleration during his 0.8 sec reaction time so a=0, from equation (A),

$S = 20 \ m/s (0.8 \ s) + 0 \times t= 16 \ m$

During the deceleration (a = -7.0 m/s^2) ,  final velocity will be zero i.e v = 0, therefore from equation (B),

$0 = (20 m/s)^2 +2 (-7.0 m/s^2) \ s \\\\ s = \frac{-400}{-14} = 28. 57 m$

Thus, the distance traveled by car after he sees need to put brakes is

$S + s = 16 \ m + 28.57 \ m =44 .57 \ m$