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).
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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