A race driver has made a pit stop to refuel. Afterrefueling, he leaves the pit area with an acceleration whosemagnitude is 6.0m/s^{2} ; after 4.0s he

Bergen

Bergen

Answered question

2021-01-15

A race driver has made a pit stop to refuel. Afterrefueling, he leaves the pit area with an acceleration whosemagnitude is 6.0ms2 ; after 4.0s he enters the mainspeedway. At the same instant, another car on thespeedway and traveling at a constant speed of 70.0m/sovertakes and passes the entering car. If the entering carmaintains it acceleration, how much time is required for it tocatch the other car? 
I tried to find the V for the car in the pit area. Isthat the right first step; what do I do next? Any help would begreat.

Answer & Explanation

Faiza Fuller

Faiza Fuller

Skilled2021-01-16Added 108 answers

Given that Intial velocity of the first car is U=0m/s.since it is starting from rest.
Its accelration is a=6m/s2. After 4seconds time it is entering in the speedway and joins with secondcar. At that time first car velocity is v=u+at
v=0+6×4
v=24m/s.
After that first car is maintaining same acceleration and secondcar is maintaining constant velocity 70m/s.
Once again after certain time first car is catching the secondmeans at that instant distance travelled by both the cars isequal.i.e., S1=S2.
First car intial velocity is 24m/s. whenjoining with second car and time of travel is same forboth.
Therefore by using equation S=ut+(12) at2.
we can write 24×t+(12×6×t2=70×t)
Since second car acceleration iszero.
Therefore 24×t+0.5×6×t2=70×t by solving thisequation we get t=15.33s.
That is first car once again catches the secondcar after 15.33s.after entering into thespeedway.

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