Question

If an object with uniform acceleration (or deceleration) has a speed of `4 m/s` at `t=0` and moves a total of 45 m by `t=3`, what was the object's rate of acceleration?

Answer 1

`a=22/3m/s^2`

Explanation:

Data:-
Initial velocity`=v_i=4m/s`
Distance `=S=45 m`
Time taken `=t=t_2-t_1=3-0=3s`
Acceleration`=a=??`
Sol:-
We know that:-
`S=v_i*t+1/2at^2`
`implies 45=4*3+1/2a3^2`
`implies 45=12+9/2a`
`implies 9/2a=33 implies 9a=66 implies a=66/9=22/3 m/s^2`
`implies a=22/3m/s^2`