Chapter A.6, Problem 99E

a.

To determine

To calculate: Rock’s velocity and acceleration as functions of time.

Answer to Problem 99E

The correct answer is velocity v (t) = $\frac{ds}{dt}=24-1.6t$ , acceleration $a(t)=\frac{d^{2}s}{d^{2}t}=-1.6$ .

Explanation of Solution

Given information: A rock is thrown vertically upward, and its initial velocity $24$ m/sec is given. To reach height in given time (t) in the function of time is also given, $s=24t-0.8t^2$ .

Formula used: We know that, Instantaneous velocity is $V(t)=\frac{ds}{dt}$ .Instantaneous acceleration is $a(t)=\frac{d^{2}s}{d^{2}t}$ . And for time to reach maximum height when $V(t)=\frac{ds}{dt}=0$ .

Calculation: By formula we can find instantaneous velocity and instantaneous acceleration,

  $\begin{array}{l}s=24t-0.8t^2\\ v(t)=\frac{ds}{dt}=24-1.6t\\ a(t)=\frac{d^{2}s}{d^{2}t}=-1.6\end{array}$

Thus, the correct answer is (a) velocity v (t) = $\frac{ds}{dt}=24-1.6t$ , acceleration $a(t)=\frac{d^{2}s}{d^{2}t}=-1.6$ .

b.

To determine

To calculate: The time to take the rock to reach its highest point.

Answer to Problem 99E

The correct answer is $15$ Sec.

Explanation of Solution

Given information: A rock is thrown vertically upward, and its initial velocity $24$ m/sec is given. To reach height in given time (t) in the function of time is also given, $s=24t-0.8t^2$ .

Formula used: We know that, Instantaneous velocity is $V(t)=\frac{ds}{dt}$ .Instantaneous acceleration is $a(t)=\frac{d^{2}s}{d^{2}t}$ . And for time to reach maximum height when $V(t)=\frac{ds}{dt}=0$ .

Calculation: By formula we can find instantaneous velocity and instantaneous acceleration,

  $\begin{array}{l}s=24t-0.8t^2\\ v(t)=\frac{ds}{dt}=24-1.6t\\ a(t)=\frac{d^{2}s}{d^{2}t}=-1.6\end{array}$

Time to reach maximum height when $V(t)=\frac{ds}{dt}=0$ .

  $\begin{array}{l}v(t)=\frac{ds}{dt}=24-1.6t=0\\ t=\frac{24}{1.6}=15\text{ sec}\end{array}$

Thus, the correct answer is $15$ Sec.

c.

To determine

To calculate: Maximum height to go rock.

Answer to Problem 99E

The correct answer is $180$ m.

Explanation of Solution

Given information: In the question, a rock is thrown vertically upward, and its initial velocity $24$ m/sec is given. To reach height in given time (t) in the function of time is also given, $s=24t-0.8t^2$ .

Formula used: We know that, Instantaneous velocity is $V(t)=\frac{ds}{dt}$ .Instantaneous acceleration is $a(t)=\frac{d^{2}s}{d^{2}t}$ . And for time to reach maximum height when $V(t)=\frac{ds}{dt}=0$ .

Calculation: By formula we can find instantaneous velocity and instantaneous acceleration,

  $\begin{array}{l}s=24t-0.8t^2\\ v(t)=\frac{ds}{dt}=24-1.6t\\ a(t)=\frac{d^{2}s}{d^{2}t}=-1.6\end{array}$

For maximum height, put $t=15$ in given equation of $s=24t-0.8t^2$ .

  $\begin{array}{l}s=24t-0.8t^2\\ $ s=24*15-0.8{(15)}^2$s=360-180\\ s=180m\end{array}$

Thus, the correct answer is $180$ m.

d.

To determine

To calculate: Time to take the rock reach half of its maximum height.

Answer to Problem 99E

The correct answer is 4.394 sec.

Explanation of Solution

Given information: In the question, a rock is thrown vertically upward, and its initial velocity $24$ m/sec is given. To reach height in given time (t) in the function of time is also given, $s=24t-0.8t^2$ .

Formula used: We know that, Instantaneous velocity is $V(t)=\frac{ds}{dt}$ .Instantaneous acceleration is $a(t)=\frac{d^{2}s}{d^{2}t}$ . And for time to reach maximum height when $V(t)=\frac{ds}{dt}=0$ .

Calculation: For time to take rock reach half of its maximum distance put $s=\frac{180}{2}=90\text{ m}$ .

  $\begin{array}{l}90=24t-0.8t^2\\ 0.8t^2-24t+90=0\\ 4t^2-120t+450=0\\ t=\frac{120\pm \sqrt{{(120)}^2-4*4*450}}{2*4}\\ t=4.394,25.60\end{array}$

But $t=25.60$ is not possible because time to reach rock to maximum height is $t=15\text{ sec}$ .Hence $t=4.394\text{ sec}$ is correct.

Thus, the correct answer is4.394 sec.

e.

To determine

To calculate: The time to the rock in the air.

Answer to Problem 99E

Explanation of Solution

Given information: In the question, a rock is thrown vertically upward, and its initial velocity $24$ m/sec is given. To reach height in given time (t) in the function of time is also given, $s=24t-0.8t^2$ .

Formula used: We know that, Instantaneous velocity is $V(t)=\frac{ds}{dt}$ .Instantaneous acceleration is $a(t)=\frac{d^{2}s}{d^{2}t}$ . And for time to reach maximum height when $V(t)=\frac{ds}{dt}=0$ .

Calculation: By formula we can find instantaneous velocity and instantaneous acceleration,

  $\begin{array}{l}s=24t-0.8t^2\\ v(t)=\frac{ds}{dt}=24-1.6t\\ a(t)=\frac{d^{2}s}{d^{2}t}=-1.6\end{array}$

Time to reach maximum height when $V(t)=\frac{ds}{dt}=0$ .

  $\begin{array}{l}v(t)=\frac{ds}{dt}=24-1.6t=0\\ t=\frac{24}{1.6}=15\text{ sec}\end{array}$

(c) For maximum height, put $t=15$ in given equation of $s=24t-0.8t^2$ .

  $\begin{array}{l}s=24t-0.8t^2\\ $ s=24\times 15-0.8{(15)}^2$s=360-180\\ s=180\end{array}$

The time to rock in the air means time to reach to get maximum height and then fall to the ground, and this is equal to two times of time taken to reach maximum height.

  $T=2\times 15=30\text{ sec}$

Thus, the correct answer 30 sec.