Question

An object, moving with constant acceleration and zero initial velocity, travels

48 m in 5.2 s. What is the magnitude of its acceleration?
3.6 m/s2
4.6 m/s2
5.6 m/s2
6.6 m/s2

Answer 1

`a=3.6(m)/(s^2)`

Step-by-step explanation:

First, use kinematic equation #3:
`\Delta x=v_it+(1)/(2)at^2`. For this problem, let

`\Delta x=48m\\v_i=0.0(m)/(s)\\t=5.2s`

Next, rewrite the kinematic equation to solve for acceleration (a):

`a=2((\Delta x-v_it)/(t^2))`

Last, put in the given values on the right side of the equation for acceleration and solve:

`a=2(((48m)-[0.0(m)/(s)(5.2s)])/((5.2s)^2))\\a=2((48m)/(27.04s^2))\\a=3.55(m)/(s^2)`