Q. 67 - Calculus

Question

A figurine that is part of the display in a mechanical music box moves back and forth along a straight track as shown next at the left. The mechanics of the music box are programmed to move the figurine back and forth with velocity 

$v\left(t\right)=0.00015t(t-5)(t-10)(t-15)(t-20)$

inches per second over the 20 seconds that the music program plays; see the graph next at the right 

(a) Use the graph just shown to describe the direction of motion of the figurine during various parts of the music program. 

(b) Determine the net distance travelled by the figurine in the first 5 seconds, the first 10 seconds, and the full 20-second music program. 

(c) Determine the total amount of distance travelled by the figurine in either direction along the track. 

Step-by-Step Solution

Answer

(a)  The "positive" direction is right and the "negative" direction is left, and then the figurine moves right along the track for the first 5 seconds then to the left for the next five seconds and continues. 

(b) The net distance travelled in 5 seconds is $\approx 5.27 inches$

The net distance travelled in 10 seconds is $\approx 3.125 inches$

The net distance travelled in 20 seconds is $\approx 0 inches$

(c) The total amount of distance travelled by the figurine in either direction along the track is $14.836$inches.

Step 1. Given information

The velocity of the machine is,

$v\left(t\right)=0.00015t(t-5)(t-10)(t-15)(t-20)$

The mechanics of the music box are programmed to move a figurine back and forth with the velocity $v\left(t\right)$inches per second over 20 seconds.

Step 2. (a) Use the graph just shown to describe the direction of motion of the figurine during various parts of the music program.

The objective is to describe the direction of motion of the figurine using the graph shown in the book during various parts of the music program.
Considering that the "positive" direction is right and the "negative" direction is left, and then the figurine moves right along the track for the first 5 seconds then to the left for the next five seconds and continues. 

Step 3. (b) Determine the net distance traveled by the figurine in the first   5 seconds, the firstsrc="data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHhtbG5zOndycz0iaHR0cDovL3d3dy53aXJpcy5jb20veG1sL21hdGhtbC1leHRlbnNpb24iIGhlaWdodD0iMjAiIHdpZHRoPSIyMyIgd3JzOmJhc2VsaW5lPSIxNiI+PCEtLU1hdGhNTDogPG1hdGggeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzE5OTgvTWF0aC9NYXRoTUwiPjxtbz4mI3hBMDs8L21vPjxtbj4xMDwvbW4+PC9tYXRoPi0tPjxkZWZzPjxzdHlsZSB0eXBlPSJ0ZXh0L2NzcyIvPjwvZGVmcz48dGV4dCBmb250LWZhbWlseT0iQXJpYWwiIGZvbnQtc2l6ZT0iMTYiIHRleHQtYW5jaG9yPSJtaWRkbGUiIHg9IjEzLjUiIHk9IjE2Ij4xMDwvdGV4dD48L3N2Zz4=" localid="1654100515945"   10 seconds, and the fullsrc="data:image/svg+xml;base64,PHN2ZyB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciIHhtbG5zOndycz0iaHR0cDovL3d3dy53aXJpcy5jb20veG1sL21hdGhtbC1leHRlbnNpb24iIGhlaWdodD0iMjAiIHdpZHRoPSIyMyIgd3JzOmJhc2VsaW5lPSIxNiI+PCEtLU1hdGhNTDogPG1hdGggeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzE5OTgvTWF0aC9NYXRoTUwiPjxtbz4mI3hBMDs8L21vPjxtbj4yMDwvbW4+PC9tYXRoPi0tPjxkZWZzPjxzdHlsZSB0eXBlPSJ0ZXh0L2NzcyIvPjwvZGVmcz48dGV4dCBmb250LWZhbWlseT0iQXJpYWwiIGZvbnQtc2l6ZT0iMTYiIHRleHQtYW5jaG9yPSJtaWRkbGUiIHg9IjEzLjUiIHk9IjE2Ij4yMDwvdGV4dD48L3N2Zz4=" localid="1654100511023"   20 -second music program.

The objective is to determine the net distance travelled by the figurine in the first $5$seconds, $10$seconds, and the full$20$ seconds music program.
The net distance travelled in $5$seconds is, 

$$\begin{gathered} {\int }_0^{5}v\left(t\right)dt={\int }_0^{5}0.00015(t-5)(t-10)(t-15)(t-20)dt \\ \approx 5.27 \end{gathered}$$

The net distance travelled in $10$seconds is,

$$\begin{gathered} {\int }_0^{10}v\left(t\right)dt={\int }_0^{10}0.00015(t-5)(t-10)(t-15)(t-20)dt \\ \approx 3.125 \end{gathered}$$

The net distance travelled in $20$seconds is,

$$\begin{gathered} {\int }_0^{20}v\left(t\right)dt={\int }_0^{20}0.00015(t-5)(t-10)(t-15)(t-20)dt \\ \approx 0 \end{gathered}$$

Step 4. (c) Finding the total amount of distance travelled by the figurine in either direction along the track.

The total amount of distance travelled by the figurine in either direction along the track. 

$$\begin{gathered} {\int }_0^5 v\left(t\right)dt={\int }_0^5 0.00015(t-5)(t-10)(t-15)(t-20)dt=5.27 \\ {\int }_5^{10} v\left(t\right)dt={\int }_0^5 0.00015(t-5)(t-10)(t-15)(t-20)dt=-2.148 \\ {\int }_{10}^{15} v\left(t\right)dt={\int }_0^5 0.00015(t-5)(t-10)(t-15)(t-20)dt=2.128 \\ {\int }_{15}^{20} v\left(t\right)dt={\int }_0^5 0.00015(t-5)(t-10)(t-15)(t-20)dt=-5.27 \end{gathered}$$

total distance D is:

$$\begin{gathered} D= 5.27 + 2.148 + 5.27 + 2.148 \\ D= 14.836 inches \end{gathered}$$