Question

Using the derivative, when is the acceleration at t=1 second?

Answer 1

The given position function is:

`s(t)=t^3-5t^2+3t`

The derivative of the position function gives the velocity function:

`v(t)=s^(\prime)(t)`

Find the derivative of the position vector:

`\begin{gathered} s^(\prime)(t)=3t^(3-1)-2\cdot5t^(2-1)+3t^(1-1) \\ \Rightarrow s^(\prime)(t)=3t^2-10t+3 \end{gathered}`

It follows that:

`v(t)=3t^2-10t+3`

The derivative of the velocity function is the acceleration function:

`a(t)=v^(\prime)(t)`

Find the derivative of the velocity function:

`\begin{gathered} v^(\prime)(t)=3\cdot2t^(2-1)-10t^(1-1)+0 \\ \Rightarrow v^(\prime)(t)=6t-10 \end{gathered}`

It implies that the acceleration is:

`a(t)=6t-10`

Substitute t=1 into the acceleration function:

`a(1)=6(1)-10=6-10=-4\text{ m/s}^2`

Hence, the acceleration at t=1 is -4m/s². ( the negative sign denotes a decrease in velocity)