Question

Your friend is on a quad is moving at 14.0 m/s when you breeze by on your bike. Your friend accelerates at 2.0 m/s2 for 3.0 seconds. How far does she travel during this time?

Answer 1

We have that the formula for the distance is:

$d=v_1t+\frac{1}{2}at^2$

Given the information in the problem, we have the following:

$\begin{gathered} v_1=14\frac{m}{s} \\ a=2\frac{m}{s^2}^{} \\ t=3s \end{gathered}$

then, applying the formula we get the following:

$\begin{gathered} d=(14\frac{m}{s})(3s)+\frac{1}{2}(2\frac{m}{s^2})(3s)^2 \\ =42m+\frac{18}{2}m=42m+9m=51m| \\ d=51m \end{gathered}$

therefore, your friend traveled 51m