# Motorcycle Catches a Car (Acceleration)

A motorcycle is 27 m behind a car that is traveling at constant speed on a straightaway. Initially, the car and the motorcycle are both traveling at the same speed, . At some time , the motorcycle starts to accelerate at a rate of . By the time the motorcycle catches up with the car, at time , it is traveling at twice the car's speed.

I solved all of the parts except for the last one

How far did the motorcycle travel from the moment when it starts to accelerate (at time ) until it catches up with the car (at time )?

The problem is I don't know how to start it

I found that Vcar= 18m/s and t2-tl=3s

and so Vmoto=36m/s

but i don't know what is wrong....

please help I have been working on this problem for about 4 hours.....I think I'm gonna burst andrevdh
Homework Helper
For the motorcycle we have from v = u + at that

$$t_c = \frac{v_o}{a}$$

where tc is the time to close the gap and vo is the intial speed of the two. Then the distance the motorcycle travelled until it caught up with the car is given by

$$v_ot_c + 0.5a{t_c}^2 = 27 + v_ot_c$$

the term on the right is the gap plus the extra distance the car travelled during the time tc. This gives

$$\frac{{v_o}^2}{a} = 54$$

using the previous formula for tc. Now the distance the motorcycle travelled can be found from

$$v^2 = u^2 + 2as$$

Last edited:
Awesome thank you