Chapter 2.5, Problem 10QR

To determine

To calculate: The acceleration of the particle, when $t=3$

Answer to Problem 10QR

The acceleration of the particle is $12m/s^2$

Explanation of Solution

Given information:

The velocity of the particle is $v=2t^3-7t^2+10$ .

Concept used:

The formula used is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ , $a=\frac{dv}{dt}$ where $a$ is the acceleration of the particle.

Calculation:

The velocity of the particle is $v=2t^3-7t^2+10$ .

Differentiate the function with respect to $t$ .

  $\begin{array}{c}\frac{d}{dt}\left(2t^3-7t^2+10\right)=\frac{d}{dt}\left(2t^3\right)-\frac{d}{dt}\left(7t^2\right)+\frac{d}{dt}\left(10\right)\\ =2\times 3t^{3-1}-7\times 2t^{2-1}+0\\ =6t^2-14t+0\\ a=6t^2-14t\end{array}$

Substitute $3$ for $t$ in the acceleration $a=6t^2-14t$ .

  $\begin{array}{c}a=6{\left(3\right)}^2-14\left(3\right)\\ =6\times 9-42\\ =54-42\\ =12\end{array}$

Conclusion: The acceleration of the particle is $12m/s^2$