Motorcycle Catches a Car-Kinematics Question

[Chandasouk]

Homework Statement

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 47.0 mph, and the distance between them is 60.0 meters. After t1 = 4.00 seconds, the motorcycle starts to accelerate at a rate of 5.00 m/s^2. The motorcycle catches up with the car at some time t2.

How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t2-t1.


The attempt at a solution

Okay, I said that the speed of the car is zero and the initial velocity of the motorcycle was zero as well. This was due to the fact that both were traveling at 47mph to begin with and in their frame of reference, they would be still.

I use the formula of

D = Xi + Vit + 1/2at^2

D meaning the distance, Xi being initial position, Vi being initial veloctiy, and a and t being acceleration and time, respectively.

60m= 0 + (o)t + 1/2(5.00m/s^2)*t^2

60m=2.5m/s^2*t^2

60m/2.5m/s^2 = t^2

24sec = t^2

4.9=t2

So, for t2 i got 4.9 seconds. Now the problem says to 4.9-4.0 which should give me .90 seconds, but when i enter that in for my answer on Mastering Physics, it says I am incorrect.

Can someone help?

 

[rl.bhat]

You have started the problem with zero relative velocity of both.
Hence t2 itself is the answer. You need not consider t1.