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:
$S(t)=S_0 + v_0 t + \frac{1}{2}at^2$
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:
$S=v_0t+ \frac{1}{2}at^2=(5.82 m/s)(3.25 s)+ \frac{1}{2}(2.35 m/s^2)(3.25 s)^2=31.3 m$