Question

Car B is traveling a distance d ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the brakes, causing his car to decelerate at 11 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 18 ft/s^2. Determine the minimum distance d between the cars so as to avoid a collision.

Answer 1

Suppose at time t both cars collide after car B applies brakes, then minimum distance between both cars is 'D'

distance traveled by car A in this time is x1

distance traveled by car B in this time is x2

x2 - x1 = D

Now Using 2nd kinematic equation:

x1 = U1*t1 + (1/2)*a1*t1^2

t1 = time for which car traveled = (t - 0.75) sec

U1 = initial speed of car A = 60 ft/sec

a1 = de-acceleration of car A = -18 ft/sec^2

x1 = 60*(t - 0.75) + (1/2)*(-18)*(t - 0.75)^2

Now for car B.

Using 2nd kinematic equation:

x2 = U2*t2 + (1/2)*a2*t2^2

t2 = time for which car traveled = t sec

U2 = initial speed of car B = 60 ft/sec

a2 = de-acceleration of car B = -11 ft/sec^2

x2 = 60*t + (1/2)*(-11)*t^2

Now

x2 - x1 = D

D = 60*t - (1/2)*11*t^2 - [60*(t - 0.75) - (1/2)*18*(t - 0.75)^2]

Now for minimum distance

dD/dt = 0

dD/dt = d[60*t - (1/2)*11*t^2 - 60*(t - 0.75) + (1/2)*18*(t - 0.75)^2]/dt = 0

dD/dt = 60 - 2*(1/2)*11*t - 60*1 + 2*(1/2)*18*(t - 0.75) = 0

18*(t - 0.75) - 11*t = 0

7*t = 18*0.75

t = 18*0.75/7 = 1.9286 sec

Now distance D will be:

D = 60*1.9286 - (1/2)*11*1.9286^2 - [60*(1.9286 - 0.75) - (1/2)*18*(1.9286 - 0.75)^2]

D = 37.0446 ft

In three significant figures

D = 37.0 ft = minimum distance between both cars

Let me know if you've any query.