# How much time elapses before the motorcycle overtakes the car?

• cowgiljl

#### cowgiljl

) A car and a mortocycle start at rest and at the same time on a straight track, but the motorcycle is 25.0 meters behind the car. The car accelrates at a uniform rate of 3.70 m/sec^2. And the motorcyle accelrates at a uniform rate of 4.40 m/sec^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 sec. later assuming that they continue at the same rate?

What i have so far

Motorcycle and the car i have set up a proportion
2x + 50/4.40 = 2x/3.70
and X is the distance which I got 132meters

MC 132 +25 = 157 meters
car = 132

using the formula d = Vi(t)
mc i got 35.7 sec using the 157m
the car was also 132meters

Where do i go from there i know i am missing something so simple

joe

car I got 35.7sec

Alright! Proper use of capitals

Alright! Proper use of capitals AND showing your work! I'm impressed- your're a fast learner!

"Motorcycle and the car i have set up a proportion
2x + 50/4.40 = 2x/3.70
and X is the distance which I got 132meters"

Do you mean "(2x+ 50)/4.40"?

Other than that it's okay. It would be better to actually state:
"After t seconds, the car will have gone x= (1/2)(3.70)t^2 meters so it will have gone x meters when t2= 2x/3.70.
After t seconds, the motorcycle will have gone y= (1/2)(4.40)t^2 meters. In order to have caught up with the car it will have to have gone y= x+25 meters. It will have gone x+ 25 meters when t2= 2(x+25)/4.40 so we must have
2(x+25)/4.40= (2x+50)/4.40= 2x/3.70.

Writing EVERYTHING out not only helps people like me follow your reasoning, it is very likely to have a good effect on your teacher and might even help you catch any mistakes. It helps me avoid losing track of what I'm doing in the middle of the problem (Something I am very prone to doing)!

Of course, that equation is the same as 3.7(2x+50)= 4.4(2x) or
7.4x+ 185= 8.8x. 1.4x= 18.5 gives x= 132 meters approximately, just as you have.

(A), however, asked for the time: Since x= 132= (1/2)(3.70)t2, this is t2= [sqrt](2(132)/3.7)= 8.44 seconds.

For (B), yes, you are correct. x= 132 meters is the distance for the car and the motorcycle traveled an additional 25 meters:
132+ 25= 157 meters.

Now the problem asks how far ahead the motorcycle will be "2 sec. later". One way to do ths would be to say: 2 seconds later than 8.44 seconds will be 10.44 seconds. Put that value of t into your formulas for the distance traveled by the car and motorcycle and subtract (remembering to allow for the car's 25 meter head start): In 10.44 seconds the car will have traveled (3.7/2)(10.44)2= 202 meters. In 10.44 seconds, the motorcyle will have (4.4/2)(10.44)2= 240 meters: that's a difference of 38 meters but remember that the motorcycle started 25 meters behind the car: it will now be 38-25= 13 meters in front. (check my arithmetic! I did this a different way and got 20 meters!)