Car B is travelling a distance d ahead of car A Both cars ar

Car B is travelling a distance d ahead of car A. Both cars are travelling at 60 ft/s when the driver of B suddenly applies the brakes, causing his car to decelerate at 9 ft/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 20 ft/s^2 Determine the minimum distance d between the cars so as to avoid a collision. Express your answer to three significant figures and include the appropriate units.

Solution

For first 0.75 sec,

Distance traveled by car A,

dA =60 x 0.75 = 45 feet

by car B,

dB = 60(0.75) - 9(0.75^2)/2 = 42.46875 feet

and speed of car B after 0.75 sec

v = 60 - 9(0.75) = 53.25 sec

Initial relative velocity of A wrt B, u = 60 - 53.25 = 6.75 feet/s

acc of A wrt to B, a = -20 + 9 = -11 ft/s^2

time taken when relative velocity becomes zero.

0 = 6.75 - 11t

t = 0.614 sec

Relative distance traveled in this time,

d1 = 6.75(0.614) - 11(0.614^2)/2 = 2.07 ft

SO initial distance d required, d = d1 + (dA - dB)

= 2.07 + ( 45 - 42.46875)= 4.605 ft ......Ans