Question

A car is moving 5.82 m/s when it accelerates at 2.35m/s^2 for 3.25s. how far does it travel?

Answer 1

The distance covered by an object moving by uniformly accelerated motion is given by:

`S(t)=S_0 + v_0 t + (1)/(2)at^2`
where

`S_0` is the intiial position

`v_0` is the initial velocity (in this case,
`v_0=5.82 m/s`)

`a` is the acceleration (in this case,
`a=2.35 m/s^2`)

We can assume
`S_0=0` since we are only interested in the distance covered by the car with respect to its initial position; so, if we substitute t=3.25 s in the equation, we can find how far the car traveled:

`S=v_0t+ (1)/(2)at^2=(5.82 m/s)(3.25 s)+ (1)/(2)(2.35 m/s^2)(3.25 s)^2=31.3 m`