Chapter 11.2, Problem 48E

Solution

a)

To find: The graph of the velocity function.

The graph of the velocity function is shown in Figure (1).

Given information:

The velocity function of the Rock is $v\left(t\right)=170-32t$ ft/sec

Calculation:

Since the function is linear. Thus, the graph of $v\left(t\right)$ is a straight line. Plot it using arbitrary points on the graph.

$t$	0	$5.313$
$v\left(t\right)=170-32t$	170	0

Thus, the graph passes through the points $\left(5.313,0\right)$ and $\left(0,170\right)$ .

The graph of the velocity function is shown in Figure (1).

  

Figure (1)

Therefore, the graph of the velocity function is shown above.

b)

To find: The time is taken by a rock to reach the maximum height.

In $5.3$ sec, the rock reaches the maximum height.

Given information:

The velocity function of the Rock is $v\left(t\right)=170-32t$ ft/sec

Calculation:

At $t=0$ , the velocity of the rock is 170 ft/sec. At the maximum height, the velocity of the rock is 0. That is $v\left(t\right)=0$ . So, to find the time taken by a rock to reach the maximum height, solve the equation $v\left(t\right)=0$ . Thus,

  $\begin{array}{c}v\left(t\right)=0\\ 170-32t=0\\ t=\frac{170}{32}\\ t=5.313\end{array}$

Hence, in $5.3$ sec, the rock reaches the maximum height.

c)

To find: The distance covered by the rock at its maximum height during the motion.

The maximum height reached by a rock is $450.5$ ft in 1.5 sec.

Given information:

The velocity function of the Rock is $v\left(t\right)=170-32t$ ft/sec

Formula used:

Area of the triangle $\frac{1}{2}\times b\times h$ .

Calculation:

The distance traveled by a rock is the same as the area under the velocity graph $v\left(t\right)=170-32t$ ft/sec. The graph is shown in Figure (1). Since the region is triangular, the area under the curve $v\left(t\right)=170-32t$ is $\frac{1}{2}\left(5.3\right)\left(170\right)\approx 451$ .

Thus, the maximum height reached by a rock is approximately $451$ ft in $5.3$ sec.