Chapter 3.4, Problem 13E

To determine

To calculate: Find (a) Rock’s velocity and acceleration as functions of time?

(b) Time to take the rock to reach its highest point?

(c) Maximum height to go rock?

(d) Time to take the rock reach half of its maximum height.

(e) The time to the rock in the air.

Answer to Problem 13E

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) $15$ Sec.

(c) $180$ m

(d) 4.394 sec.

(e) 30 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: By formula we can find instantaneous velocity and instantaneous acceleration,

(a)

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

(b) 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*15-0.8{(15)}^2$s=360-180\\ s=180m\end{array}$

(d) 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.

(e) 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.

  $T=2*15=30\text{ sec}$

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) $15$ Sec.

(c) $180$ m

(d) 4.394 sec.

(e) 30 sec.