# A car moving with

Let's put kinematics to play, shall we?

Consider the third equation of motion,

v2−u2 =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).

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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