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Question

# A car starts from rest and accelerates at 1 ms−2 for 4 s. It then moves at a constant velocity for next 10 s.then decelerates at 2.0 ms−2 adn finally comes to rest.The total distance travelled by the car is

A
52 m
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B
40 m
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C
48 m
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D
46 m
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Solution

## The correct option is A 52 mDuring first 4s the distance traveled, x1=ut+at2/2=0+1×162=8mAt the end of these 4s it will gain velocity v=at=1×4=4m/sSo the new distance in next 10sec with constant velocity v is x2=v×10=40mand then it comes to rest after travelling a distance s with retardation a1 so this s=v22a1=162×2=4mTotal distance is d=x1+x2+s=8+40+4=52mOption A is correct.

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