Question

Car B is traveling a distance d ahead of car A. Both cars are traveling at 60 m/s when the
driver of B suddenly applies the brakes, causing his car to decelerate at 12 m/s 2 . It takes
the driver of car A 0.75 s to react (this is the normal reaction time for drivers). When he
applies his brakes, he decelerates at 15 m/s 2 . Determine the minimum distance d between
the cars so as to avoid a collision.

Answer 1

Initially the relative speed of both the cars with respect to each other were Zero.

Now when car B applied breaks, the relative acceleration of car A wrt car B becomes 12 m/s² towards car B. And for 0.75 s the condition was same.

Now using the equation v-u=at
we can find the relative speed of car A wrt car B after 0.75s
So, v= (12*0.75) + 0
v = 9 m/s
So that means the distance reduced in between the two cars in 0.75s = 9x0.75 = 6.75m


Now after 0.75s car A also applied breaks that means the relative acceleration of car A wrt car B reduced to 12-15 = 3m/s² but in reverse direction.

Now the time taken by the car A to reduce it's relative speed from 9m/s to 0m/s with deceleration of 3m/s² will be
v= u+at
t=(v-u)/a=(0-9)/(-3)=3s

Now distance travelled with this relative deceleration of 3m/s² of car A wrt car B while car A is going toward car B with relative speed of 9m/s, in 3s will be
s= ut+0.5at²
s=(9x3) + (0.5x(-3)x3²)
=27+(-27/2)
=27/2
= 13.5 m

So the minimum distance required to avoid Collision will be = 6.75+13.5 = 20.25m.