Q. 14.7( 3 Votes )

# A car moving with an initial velocity of 10 m/s slows down at a constant deceleration and stops after traveling 400 m. What is the time (in seconds) taken by the car to come to a stop?

A. 80 sec

B. 60 sec

C. 25 sec

D. 40 sec

Answer :

Let's put kinematics to play, shall we?

Consider the third equation of motion,

v^{2}−u^{2} =2×a×s ------- (1)

Here , 'v' is the final velocity of the body in question

' u ' is the initial velocity.

' a ' is the acceleration experienced.

' s ' is the distance traveled.

Taking only S.I. units, for the case mentioned in the question,

it can be calculated that a = -1/8 ms^{-2} (v = 0, u = 10, s = 400).

The negative sign indicates that the body (the car) is decelerating.

Applying this result to first equation of motion,

v=u + a×t -------- (2)

You can calculate the time taken (t) as t = -10/(-1/8).

Which is:t = 80 seconds.

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