A car is traveling at a constant speed of 25.5 m/s on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 2.35 km away?

Answer 1

d = V*t = 2350 m.
25.5*t = 2350.
t = 92.2 s. To reach the next exit.

d = 0.5a*t^2 = 2350 m.
0.5a*92.2^2 = 2350.
4250.42a = 2350.
a = 0.553 m/s^2.