Question

In the attached figure, car B advances at 72.0 km / h when it begins to brake with a constant acceleration aB = -6.00 m / s2. The car A behind, at a distance d = 20.0 m from the car B, also advances to 72.0 km / h and when seeing that the car B brakes, it also brakes with a constant acceleration aA = -4.00 m / s2. However, the driver of car A has a reaction time of t1 = 0.7 s (since he sees that car B starts to brake, until it actually starts to brake). Determine the time t * (in s) that the cars take to crash, measured from the moment the car A begins to brake. Write your result in the attached box, without units and using 3 significant figures.

Answer 1

speed of B=20 m/s; speed of A=20 m/s

Let distance travelled by car B is 'd' metres while braking, let the time taken to cover this distance is (t+0.7) seconds

distance covered by A in constant velocity, = 0.7*20 = 14 m

distance travelled by A while decelerating, s = d+(20-14) = (d+6) metres

time taken by the car A to decelerate is t

For car B:

d=20*(t+0.7) - 0.5*6*(t+0.7)²

d= 20t + 14 - 3t² -1.47 + 4.2t

For car A:

d+6 = 20*t - 0.5*4*(t)²

d+6=20t - 2t²

equating the value of d from both the equations,

20t + 14 - 3t² -1.47 + 4.2t + 6 = 20t -2t²

18.53 -t² +4.2t =0

solving the above equation we get,

t = 6.8895 seconds