Question #3742

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 3.70 m/s2, and the motorcycle at a uniform rate of 4.40 m/s2. How much time elapses before the motorcycle overtakes the car? How far will each have traveled during that time? How far ahead of the car will the motorcycle be 2.00 s later?What is the distance traveled of both car and the motorcycle within that time? (Both vehicles are still accelerating.)

Expert's answer

a1=3.7a2=4.4

h=25

Let t - time elapse before the motorcycle overtakes the car then

h+a1*t^{2}/2=a2*t^{2}/2 = > t=sqrt( 2*h/(a2-a1) )=8.45 sx1=a1*t^{2}/2=132.1 my1=a2*t^{2}/2=157.1 mafter T=t+2=10.45 seconds the car will be far ahead of the motorcycle:

h1=(h+a1*T^{2}/2-a2*T^{2}/2 )=63.23 mx2=a1*T^{2}/2=202.1 my2=a2*T^{2}/2=240.3 m

h=25

Let t - time elapse before the motorcycle overtakes the car then

h+a1*t

h1=(h+a1*T

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