Chapter 11.2, Problem 47E

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 $v\left(t\right)=48-32t$ ft/sec

Calculation:

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

$t$	0	1.5
$v\left(t\right)=48-32t$	48	0

Thus, the graph passing through the points $\left(1.5,0\right)$ and $\left(0,48\right)$ .

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 required for the rock to reach the maximum height.

The rock reaches the maximum height after 1.5 seconds of the motion.

Given information:

The velocity function of the Rock is $v\left(t\right)=48-32t$ ft/sec

Calculation:

At $t=0$ , the velocity of the rock is 48 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,

  $\begin{array}{c}v\left(t\right)=0\\ 48-32t=0\\ t=\frac{48}{32}\\ t=1.5\end{array}$

Hence, in 1.5 sec, the rock reaches the maximum height.

c)

To find: The distance covered by the rock has traveled at its maximum height.

The maximum height reached by a rock is 36 ft in 1.5 sec.

Given information:

The velocity function of the Rock is $v\left(t\right)=48-32t$ ft/sec

Formula used:

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

Calculation:

The distance traveled by the rock is the same as the area under the velocity graph $v\left(t\right)=48-32t$ ft/sec. The graph is shown in Figure (1). Since, the region is rectangular, the area under the curve $v\left(t\right)=48-32t$ is $\frac{1}{2}\left(1.5\right)\left(48\right)=36$ .

Thus, the maximum height reached by a rock is 36 ft in 1.5 sec.