Chapter 11.2, Problem 8QR

To determine

To calculate: Theposition of the particle when $t=4$ .

Answer to Problem 8QR

The position of the particle at $t=4$ is 8.

Explanation of Solution

Given information:

Velocity $v\left(t\right)=3t^2-12t$ for $t\ge 0$

Position $\left(x\right)=40$ at $t=0$

Time $\left(t\right)=4$

Formula used:

Velocity of an object with distance $x(t)$ at time $t$ :

  $x(t)={\displaystyle \int v\left(t\right)}dt$

Calculation:

Given that the velocity of the particle is $v\left(t\right)=3t^2-12t$ .

Now, calculate position. Use the formula $x(t)={\displaystyle \int v\left(t\right)}dt$ .

  $\begin{array}{c}x(t)={\displaystyle \int v\left(t\right)}dt\\ ={\displaystyle \int \left(3t^2-12t\right)}dt\\ =3\frac{t^3}{3}-12\frac{t^2}{2}+C\\ =t^3-6t^2+C\end{array}$    (1)

Substitute $x=40$ and $t=0$ in the equation (1).

  $\begin{array}{c}x(t)=t^3-6t^2+C\\ 40=0^3-6\times 0^2+C\\ C=40\end{array}$

Now, calculate $x$ for $t=4$ .

  $\begin{array}{c}x(4)=4^3-6\times 4^2+40\\ =64-96+40\\ =8\end{array}$

Conclusion:

Hence, the required position of the particle at $t=4$ is 8.