Answer to Question #96565 in Mechanics | Relativity for Mohit verma

Question

Tom started chassing Jerry from the dinning table ran behind him for 16.2m toward south-west from there 2m toward the east .Then 9m toward north and then 2m toward east and finally he ran for 7m till the door in the north and caught Jerry .How far is the dinning table with respect to the door?

Answer 1

Find the Distance

We need to find, How far is the dinning table with respect to the door.

Solution:

In the 3rd Quadrant sin and cos functions are negative.

Let $\vec {D1}$ = Distance along S-W = 16.2( $- \frac {1} {\sqrt 2} \vec {i} - \frac {1} {\sqrt 2} \vec {j} )$ ( cos 45 = sin45 = $\frac {1} {\sqrt 2}$)

$\vec {D2}$ = Distance Along East = $2 \vec {i}$

$\vec {D3}$ =Distance along north = 9 $\vec {j}$

$\vec {D4}$ = Distance along east = 2 $\vec {i}$

$\vec {D5}$ = Distance along north = 7$\vec {j}$

$\vec {D}$ = $\vec {D1} + \vec {D2} + \vec {D3} + \vec {D4} + \vec {D5}$

= 16.2 ($- \frac {1} {\sqrt 2} \vec {i} - \frac {1} {\sqrt 2} \vec {j} )$ + $2 \vec {i} + 9 \vec {j} + 2 \vec {i} + 7 \vec {j}$

= - 7.3 $\vec {i} +4.6 \vec {j}$

Complete Distance Travelled = $\vert { \vec { D} } \vert$ = $\vert -7.3\vec {i} + 4.6 \vec {j} \vert$

= $\sqrt {{(-7.3)}^2 + {(4.6)}^2} = \sqrt {74.45} = 8.62 m$

Answer = 8.62 m