How Can a Motorcycle Overtake a Car with Different Accelerations?

[charitysmama]

[SOLVED] Kinematic Equations

This is the question. I have sat here and tried to figure out how to even start this problem for hours. I can't seem to get it through my head. If someone could help me set it up, I can figure out the rest I think. Thanks in Advance.

Q. A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The car accelerates at a uniform rate of 3.70m/s^2 and the motorcycle at a uniform rate of 4.40m/^2. (a) How much time elapses before the motorcycle overtakes the car? (b) How far will each have traveled during that time? (c) How far ahead of the car will the motorcycle be 2.00 s later? (Both vehicles are still accelerating.)

Anyway... Some assistance in setting it up would be much appreciated. Thanks.

 

[Hootenanny]

Welcome to PF,

Start by making a list of the things that you know and the things that you don't know.

 

[charitysmama]

I know the initial position of the motorcycle is 0m and the initial position of the car is 25m. Initial velocity of both is 0 m/s. Acceleration of the motorcycle is 4.40m/s^2 and the car is 3.70m/s^2. Initial time for both is 0 s. I do not know final position, final velocity. Final time is what I am looking for.

 

[Hootenanny]

I know the initial position of the motorcycle is 0m and the initial position of the car is 25m. Initial velocity of both is 0 m/s. Acceleration of the motorcycle is 4.40m/s^2 and the car is 3.70m/s^2. Initial time for both is 0 s. I do not know final position, final velocity. Final time is what I am looking for.

Perhaps it would be easier if you considered their relative position and accelerations. Suppose you define your reference frame as the frame of the motorcycle. Then the relative position of the car to the motorcycle is 25m and the relative acceleration of the car to the bike is $(3.70-4.40)m.s^{-2}$. Do you follow?

 

[charitysmama]

Not at all...

I'm sorry. I've done well up until these. I do not really like asking for help... but for whatever reason, this one is flying right past me. I appreciate the help.

 

[Hootenanny]

Not at all...

I'm sorry. I've done well up until these. I do not really like asking for help... but for whatever reason, this one is flying right past me. I appreciate the help.

No problem. Okay, imagine that you are sat on the bike. Initially, you measure the distance between you and the car as 25m. Then, you both begin to accelerate, the car is accelerating at 3.70 m/s^2 relative to the ground, but you are also accelerating at 4.40 m/s^2 relative to the ground. Therefore, from your point of view on the bike, the acceleration of the car is (3.70 - 4.40)m/s^2.

Does that explain it any better?

 

[charitysmama]

So... It would be (-0.7)m/s^2? Acceleration that is...

 

[Hootenanny]

So... It would be (-0.7)m/s^2? Acceleration that is...

Correct. So for question (a) you want to know the time t at which the relative position of the car to the bike is x=0 (i.e. when the bike is level with the car). You know the initial [relative] initial position and the relative acceleration, so which kinematic equation should you use?

 

[charitysmama]

Um... Xf=Xi+Vi(tf-ti)+1/2a(tf-ti)^2? Maybe?

 

[Hootenanny]

Correct

 

[charitysmama]

Thanks... got it done just in time for class... then he decided we wouldn't turn it in. Grr...