Chapter 4.7, Problem 1E

To determine

To find the velocity and acceleration at time t of a point moving on coordinate line along the position function $s(t)=3t^2-12t+1$ in the interval $[0,5]$ .

Answer to Problem 1E

  $v(t)=6t-12,a(t)=6$

Explanation of Solution

Given:

Given function: $s(t)=3t^2-12t+1$

Interval: $[0,5]$

Concept used:

To find the velocity of particle, differentiate the position function with respect to time.

To find the acceleration of a particle, differentiate the velocity function with respect to time.

Calculation:

  $s(t)=3t^2-12t+1$

Differentiate with respect to t,

  $\begin{array}{l}\Rightarrow s\text{'}(t)=3\times 2t-12\times 1\\ \Rightarrow v(t)=6t-12\end{array}$

Differentiate with respect to t,

  $\begin{array}{l}\Rightarrow v\text{'}(t)=6\times 1-0\\ \Rightarrow a(t)=6\end{array}$

Description about motion of the point:

  $\begin{array}{l}v(t)=6t-12\\ \Rightarrow v(0)=6(0)-12\\ =0-12\\ =-12\end{array}$

  $\begin{array}{l}\Rightarrow v(5)=6(5)-12\\ =30-12\\ =18\end{array}$

In the interval $[0,5],$ the velocity is changing from negative to positive.

It means, the point decreases its velocity initially and then increases.

  $a(t)=6$

In the interval $[0,5],$ the acceleration of point is constant.

Conclusion:

Therefore, the velocity of point is $v(t)=6t-12$ and the acceleration of point is $a(t)=6.$