- #1
Kalie
- 46
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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 going to burst
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 going to burst