Chapter 10.1, Problem 53E

To determine

To find time taken by the carriage to travel to midpoint of the cable

Answer to Problem 53E

Time taken by carriage to travel to midpoint of the cable $=178\text{seconds}$

Explanation of Solution

Given information : 

Speed of suspended carriage $=2.25\text{meters per second}$

Initial point of cable $\left(1,0,1\right)$

Final point of cable $\left(2,800,41\right)$

Concept used:

Midpoint of line segment joining points $\left(x_1,x_2,x_3\right)\text{ and }\left(y_1,y_2,y_3\right)$ in space $=\left(\frac{x_{{}_1}+y_{{}_1}}{2},\frac{x_2+y_2}{2},\frac{x_3+y_3}{2}\right)$

Distance between two points $\left(x_1,x_2,x_3\right)\text{ and }\left(y_1,y_2,y_3\right)$ = $\sqrt{{\left(y_1-x_1\right)}^2+{\left(y_2-x_2\right)}^2+{\left(y_3-x_3\right)}^2}$

Time= $\frac{Dis\text{tance}}{Speed}$

Explanation: 

Find midpoint of cable

Midpoint of cable

  $\begin{array}{l}=\left(\frac{1+2}{2},\frac{0+800}{2},\frac{1+41}{2}\right)\\ \text{=}\left(\frac{3}{2},\frac{800}{2},\frac{42}{2}\right)\\ \text{=}\left(\frac{3}{2},400,21\right)\end{array}$

Distance between initial point and midpoint of cable

  $\begin{array}{l}=\sqrt{{\left(\frac{3}{2}-1\right)}^2+{\left(400-0\right)}^2+{\left(21-1\right)}^2}\\ \text{=}\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(400\right)}^2+{\left(20\right)}^2}\\ =\sqrt{\frac{1}{4}+160000+400}\\ \text{=}\sqrt{\frac{641601}{4}}\\ \text{=}\frac{801}{2}\text{meters}\end{array}$

Required time

  $\begin{array}{l}=\frac{801}{2}÷2.25\\ =\frac{801}{2}\times \frac{100}{225}\\ \text{=178 seconds}\end{array}$

Hence,

A suspended carriage will take 178 seconds to travel to the midpoint of the cable