Chapter 2.4, Problem 13E

a.

To determine

To calculate: The velocity and time at time $t$

Answer to Problem 13E

The velocity is $v=24-1.6t$ and acceleration is $a=-1.6$ .

Explanation of Solution

Concept used:

Differentiate the position with respect to $t$ , get velocity and velocity with respect to $t$ to get acceleration.

Given information:

The position function is $s=24t-0.8t^2$ and velocity is $24m/s$ .

Calculation:

The function is $s=24t-0.8t^2$ .

Differentiate the position $s=24t-0.8t^2$ with respect to $t$ .

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left(24t-0.8t^2\right)\\ =\frac{d}{dt}\left(24t\right)-\frac{d}{dt}\left(0.8t^2\right)\\ =24-0.8\left(2t\right)\\ v=24-1.6t\end{array}$

The velocity is $v=24-1.6t$ .

Differentiate the velocity $v=24-1.6t$ with respect to $t$ .

  $\begin{array}{c}\frac{dv}{dt}=\frac{d}{dt}\left(24-1.6t\right)\\ =\frac{d}{dt}\left(24\right)-\frac{d}{dt}\left(1.6t\right)\\ a=-1.6\end{array}$

The acceleration is $a=-1.6$ .

Conclusion: So, the velocity is $v=24-1.6t$ and acceleration is $a=-1.6$ .

b.

To determine

To calculate: The time taken to reach the highest point of the rock.

Answer to Problem 13E

The time is $t=15\sec$ .

Explanation of Solution

Given information:

The position function is $s=24t-0.8t^2$ and velocity is $24m/s$ .

Calculation:

The function is $s=24t-0.8t^2$ .

Differentiate the position $s=24t-0.8t^2$ with respect to $t$ .

  $\begin{array}{c}\frac{ds}{dt}=\frac{d}{dt}\left(24t-0.8t^2\right)\\ =\frac{d}{dt}\left(24t\right)-\frac{d}{dt}\left(0.8t^2\right)\\ =24-0.8\left(2t\right)\\ v=24-1.6t\end{array}$

The velocity is $v=24-1.6t$ .

To get the time to reach the highest point of the rock ,equate the velocity in terms of $t$ to zero.

  $\begin{array}{l}24-1.6t=0\\ 1.6t=24\\ t=\frac{24}{1.6}\\ t=15\sec \end{array}$

Conclusion: So, the time is $t=15\sec$ .

c.

To determine

To calculate: The maximum height which the rock has reached.

Answer to Problem 13E

The maximum height is $s=180m$ .

Explanation of Solution

Concept used:

The formula is $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Given information:

The position function is $s=24t-0.8t^2$ and velocity is $24m/s$ .

Calculation:

The time taken by the rock to reach the maximum height is $t=15\sec$ .

Substitute $15$ for $t$ in the position function $s=24t-0.8t^2$ .

  $\begin{array}{c}s=24\left(15\right)-0.8{\left(15\right)}^2\\ =360-180\\ s=180m\end{array}$

Conclusion: So, the maximum height is $s=180m$ .

d.

To determine

To calculate: The time taken to cover half of maximum height reached by the rock..

Answer to Problem 13E

The time taken is $t=4.39\sec$ .

Explanation of Solution

Given information:

The position function is $s=24t-0.8t^2$ and velocity is $24m/s$ .

Concept used:

The quadratic formula is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

The maximum height is $s=180m$ .

So half of the maximum height is $90m$

Substitute $90$ for $s$ in the position function $s=24t-0.8t^2$ .

  $\begin{array}{c}90=24t-0.8t^2\\ 0.8t^2-24t+90=0\\ t=\frac{-\left(24\right)\pm \sqrt{{\left(24\right)}^2-4\times 0.8\times 90}}{2\times 0.8}\\ =\frac{-24\pm \sqrt{576-288}}{1.6}\end{array}$

Solve further.

  $\begin{array}{c}t=\frac{24\pm \sqrt{288}}{1.6}\\ =\frac{24\pm 16.97}{1.6}\\ =\frac{24+16.97}{1.6},\frac{24-16.97}{1.6}\\ t=25.6,4.39\end{array}$

The time cannot be $25.6\sec$ as the time taken to reach maximum height is only $t=15\sec$ , which is impossible.

Conclusion: So, the time is $t=4.39\sec$ .

e.

To determine

To calculate: The time when the rock was in the air.

Answer to Problem 13E

The time is $t=30\sec$ .

Explanation of Solution

Given information:

The position function is $s=24t-0.8t^2$ and velocity is $24m/s$ .

Calculation:

Equate the position function $s=24t-0.8t^2$ to zero to get the time it was in the air.

  $\begin{array}{l}24t-0.8t^2=0\\ 24t=0.8t^2\\ 0.8t=24\\ t=30\sec \end{array}$

Conclusion: So, the time the rock was in the air is $t=30\sec$ .