A car is traveling at constant speed of 27m/a on a highway.?

Question

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 1.8 km away?

Answer 1

The eqn for distance is s = v0t + 1/2a*t^2

s = distance = 1.8 km = 1800m

v0 = initial velocity (which for the first vehicle = 27m/s, for the second vehicle = 0)

a = acceleration (which for the first vehicle = 0, for the second vehicle = a)

t = time

Solving t for the first vehicle, s = 1800m = 27m/s*t + ½*0*t^2 = 27m/s*t, t = 66.67 sec

Substituting t=66.67 sec to solve for a for the second vehicle, s = 1800m = 0*t + ½*a*(66.67)^2, a = .81m/s^2