Chapter 2, Problem 72P

(a)

To determine

Whether the passenger in the elevator is gone to a higher or lower floor.

Answer to Problem 72P

The passenger in the elevator is gone to a higher floor.

Explanation of Solution

From the given graph the motion of the elevator is represented as an area under curve for the change in velocity. Each space along the t axis represents $2\text{s}$.

$A_1=\frac{1}{2}bh$ (I)

Here, $A_1$ is the area of the curve, $b$ is the breadth, and $h$ is the height.

Conclusion:

Substitute $8\text{s}$ for $b$ and $0.2{\text{m/s}}^2$ for $h$ in equation (I) for area $A_1$.

$\begin{array}{c}{A}_1=\frac{1}{2}\left(8\text{s}\right)\left(0.2{\text{m/s}}^2\right)\\ =0.8\text{m/s}\end{array}$

The elevator accelerates $\left(a_y>0\right)$ to $0.8\text{m/s}$ during the first $8\text{s}$. Then, the elevator travels at $0.8\text{m/s}$$\left(a_y=0\right)$ for next $8\text{s}$.

Substitute $4\text{s}$ for $b$ and $-0.4{\text{m/s}}^2$ for $h$ in equation (I) for area $A_2$.

$\begin{array}{c}{A}_2=\frac{1}{2}\left(4\text{s}\right)\left(-0.4{\text{m/s}}^2\right)\\ =-0.8\text{m/s}\end{array}$

The elevator slows down $\left(a_y<0\right)$ until it come to rest. Then it sits for the next $4\text{s}$. Therefore, the passenger has gone to a higher floor.

(b)

To determine

Sketch the graph of the velocity Versus time.

Answer to Problem 72P

The velocity $v_y$ of the elevator versus time is shown in the graph.

Explanation of Solution

Sketch the graph $v_y$ Versus $t$ by plotting points at two seconds interval with $v_y$ determined from the $a_y$ versus $t$. Each rectangle represents $\left(2\text{s}\right)\left(0.2{\text{m/s}}^2\right)=0.4\text{m/s}$.

$t$(s)	$v_y$	$t$(s)	$v_y$
0	The elevator is at rest, so $v_y=0$	12	$0.8\text{m/s}$
2	$1/4\left(0.4\text{m/s}\right)=0.1\text{m/s}$	14	$0.8\text{m/s}$
4	$1\left(0.4\text{m/s}\right)=0.4\text{m/s}$	16	$0.8\text{m/s}$
6	$7/4\left(0.4\text{m/s}\right)=0.7\text{m/s}$	18	$-1\left(0.4\text{m/s}\right)+0.8\text{m/s}=0.4\text{m/s}$
8	$2\left(0.4\text{m/s}\right)=0.8\text{m/s}$	20	$-1\left(0.4\text{m/s}\right)+0.4\text{m/s}=0$
10	$0.8\text{m/s}$, since $a_y=0$	22	0, since $a_y=0$

Conclusion:

The graph shows the velocity $v_y$ of the elevator versus time taken.

(c)

To determine

Sketch the graph of the position Versus time.

Answer to Problem 72P

The acceleration $a_y$ of the elevator versus time is shown in the graph.

Explanation of Solution

Each vertical space represents $0.1\text{m/s}$ and each horizontal space is $2$. It is increasing until the height is stops.

Break the region under the graph from part (b) into six sections:

Section 1: $\left(0,\text{ 4}\text{s}\right)$

Section (2):$\left(4\text{s, 8}\text{s}\right)$

Section (3):$\left(8\text{s, 16}\text{s}\right)$

Section (4):$\left(16\text{s, 18}\text{s}\right)$

Section (5):$\left(18\text{s, 20}\text{s}\right)$

Section (6):$\left(20\text{s, 24}\text{s}\right)$

The approximate distance of the elevator travels in each section by counting boxes under the curves each of them represents displacement $\left(0.1\text{m/s}\right)\left(2\text{s}\right)=0.2\text{s}$.

Section 1:$3\left(0.2\text{m}\right)=0.6\text{m}$

Section (2):$13\left(0.2\text{m}\right)=2.6\text{m}$

Section (3):$32\left(0.2\text{s}\right)=6.4\text{m}$

Section (4):$6.5\left(0.2\text{m}\right)=1.3\text{m}$

Section (5):$1.5\left(0.2\text{m}\right)=0.3\text{m}$

Section (6):$0\left(0.2\text{m}\right)=0$

The plotting data shows the position at the end of each time interval:

$t$(s)	$x$(m)
0	0
4	0.6
8	3.2
16	9.6
18	10.9
20	11.2
24	11.2

Conclusion:

The graph shows the position $x$ of the elevator versus time taken.