What is the acceleration of an object if it slows from 2 m/s to 1.3 m/s in a distance of .02 meters?

Question

Also, what is the amount of time it takes for the acceleration?

Answer 1

This question is the same as the other one you asked. Here is the answer again.

s = v0t + (1/2)at^2
v = v0 + at and t = (v - v0)/a
s = v0(v - v0)/a + (1/2)a (v - v0)^2/a^2
2as = 2vv0 - 2v0^2 + (v^2 - 2vv0 + v0^2)
2as = v^2 - v0^2
v^2 = v0^2 + 2as

v0 = 2
v = 1.3
s = 2 cm = .02 m

1.3^2 = 2^2 + 2a(.02)
a = -57.75 m/s^2 = 5.893 g's

v = v0 + at = 2 - 57.75t = 1.3
t = .01212 seconds

Answer 2

If you can assume that the acceleration is constant, you can use this formula that relates distance, acceleration and time:

d = at²/2

(Note: if acceleration is NOT constant, the above formula won't work. In such a case, you need to supply an additional formula that indicates how the acceleration changes over time.)