Question

Two cars are traveling along a straight line in the same direction, the lead car at 25.0 m/s and the other car at 28.0 m/s. At the moment the cars are 40.0 m apart, the lead driver applies the brakes, causing his car to have an acceleration of 2.00 m/s?. (a) How long does it take for the lead car to stop? (b) Assuming that the chasing car brakes at the same time as the lead car, what must be the chasing car's minimum negative acceleration so as not to hit the lead car? m/s? (c) How long does it take for the chasing car to stop?

Answer 1

The leading car have initial velocity u1 = 25 m/s

The rear car have initial velocity u2 = 28 m/s

The distance between the car is 40.0 m

The accleration of the lead car is a1 = - 2.0 m/s2

a) The time taken by the lead car to stop is final velocity v1 = 0 m/s

$t = \frac{v_1 - u_1}{a}$

putting the value we get time t = 12.5 s

b) The distance travelle by the lead car is

$s = \frac{v_1^2 - u_1^2}{2a}$

putting the value we get , displacement s = 156.25 s

the acceleration of the rear car is given by

$a_2 = \frac{v_2^2 - u_2^2}{2s}$

putting the velocity v2 = 0 m/s as the body is at rest.

putting the value we get

a2 = 2.5 m/s2

c)  The time taken by the rear car to stop is final velocity v2 = 0 m/s

$t_2 = \frac{v_2 - u_2}{a_2}$

putting the value we get time t = 11.5 s