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A race driver has made a pit stop to refuel. After refueling, he

starts from rest and leaves the pit area with an acceleration whose

magnitude is 5.9 m/s2; after 4.3 s he enters the main speedway. At

the same instant, another car on the speedway and traveling at a

constant velocity of 69.7 m/s overtakes and passes the entering

car. The entering car maintains its acceleration. How much time is

required for the entering car to catch up with the other car?

Edit
Added Mon, 11 Jul '16
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Let t be the time required catch up

and t starts from the instant the car enters the main speed

way

So, the distance travelled by car

in first 4.3 s

S = ut 1/2at2 = 0

1/2 X 5.9 X 4.32 = 54.54 m

And distance travelled by it

after (t 4.3) s

S = 0 1/2 X 5.9 (t

4.3) 2

= 2.95(t2 8.6t 18.49 )

= 2.95 t2 25.37t 54.54

So disatnce travlled in t s

S = 2.95 t2 25.37t 54.54 - 54.54

= 2.95 t2 25.37t

Distance travlled by another car in t time

S = 69.7 t

Both distance will be same to catch up

2.95 t2 25.37t = 69.7 t

t = 44.33 / 2.95 = 15.027 s

Edit
Added Mon, 11 Jul '16
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