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+(1)/(2)at^2`

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

`\begin{gathered} v_1=14(m)/(s) \\ a=2(m)/(s^2)^{} \\ t=3s \end{gathered}`

then, applying the formula we get the following:

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

therefore, your friend traveled 51m