Question

A particle moving along the x axis has a position given by x = (24t – 2.0t 3) m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero?

Answer 1

The velocity is defined by:

`v=(dx)/(dt)`

where x is the position of the particle and t is the time.

Plugging the position function given we have that the velocity is:

`\begin{gathered} v=(dx)/(dt) \\ =(d)/(dt)(24t-2t^3) \\ =24-6t^2 \end{gathered}`

Hence the velocity is given by the function:

`v=24-6t^2`

to determine the isntant when the velocity is zero we equate its expression to zero and solve for t:

`\begin{gathered} 24-6t^2=0 \\ 6t^2=24 \\ t^2=(24)/(6) \\ t^2=4 \\ t=\pm\sqrt[]{4} \\ t=\pm2 \end{gathered}`

Since time is always positive we conclude that the velocity is zero at t=2 s.

Now that we know at which instant the velocity is zero we need to remember that the acceleration is defined as:

`a=(dv)/(dt)`

then we have that:

`\begin{gathered} (dv)/(dt)=(d)/(dt)(24-6t^2) \\ =-12t \end{gathered}`

hence the acceleration is:

`a=-12t`

Plugging the value we found for the time we have that:

`a(2)=-12(2)=-24`

Therefore the acceleration of the particle when its velocity is zero is -24 meters per second per second.