Question

A car slows down uniformly from a speed of 30.0 m/s to rest in 7.20 s

Answer 1

When acceleration is constant, the average velocity is given by

$\bar v=\dfrac{v+v_0}2$

where $v$ and $v_0$ are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by

$\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t-t_0}$

where $x,x_0$ are the final/initial displacements, and $t,t_0$ are the final/initial times, respectively.

Take the car's starting position to be at $t_0=0\,\mathrm s$. Then

$\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t$

So we have

$x=0\,\mathrm m+\dfrac12\left(0\,\dfrac{\mathrm m}{\mathrm s}+30.0\,\dfrac{\mathrm m}{\mathrm s}\right)(7.20\,\mathrm s)=108\,\mathrm m$

You also could have first found the acceleration using the equation

$v=v_0+at$

then solve for $x$ via

$x=x_0+v_0t+\dfrac12at^2$

but that would have involved a bit more work, and it turns out we didn't need to know the precise value of $a$ anyway.