Question

a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance traveled

Answer 1

Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

$a=\dfrac{\Delta v}{\Delta t}=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.47\,\mathrm s}=11.2\,\dfrac{\mathrm m}{\mathrm s^2}$

Now, we have that

${v_f}^2-{v_0}^2=2a\Delta x$

so we end up with a distance traveled of

$\left(46.1\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(11.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x$

$\implies\Delta x=79.6\,\mathrm m$