If displacement is 72.1 m, initial velocity is zero and time is 0.735 s, find the acceleration.

Hello!
Probably a body moves with a uniform acceleration, denote it a. Then the speed changes uniformly, V(t) = V_0 + a t, where V_0 is the initial speed and t is the time in seconds since the initial moment. Because the initial speed is given to be zero, we have V(t) = a t.
Therefore the displacement from the initial position is equal to D(t) = (a t^2) / 2 (proving this requires integration or computing the area of a triangle but I hope you know this fact).
The unknown in our problem is a, to find it we multiply both sides of the equation (a t^2) / 2 by 2 and divide by t^2 and obtain the answer  a = (2 D) / t^2.
Numerically it is equal to  (2 * 72.1) / 0.735^2 approx 267 (m / s^2).
 
http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html

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