Question

In its daily prowl of the neighborhood, a cat makes a displacement of 118 m due north, followed by a 86 m displacement due west. If the cat takes 43 minutes to complete the 118 m displacement and 19 minutes to complete the 86 m displacement, what are the magnitude and direction of its average velocity during this 62-minute period of time

Answer 1

Answer:146.01 m

Explanation:

Given

First cat makes a displacement of 118 m towards north, Followed by a 86 m displacement due west.

Total displacement will be line joining the starting and the end point given by

$=\sqrt{118^2+86^2}=\sqrt{21,320}=146.01 m$

its direction would be given if we consider the position vector of final point

$r=-86\hat{i}+118\hat{j}$

let $\theta$is the angle made by vector with Y-axis

$tan\theta =\frac{118}{86}=1.372$

$\theta =\approx 53.91^{\circ} w.r.t to west$

and $180-53.91=126.09^{\circ} w.r.t east$

average velocity$=\frac{displacement}{time\ taken}$

$V_{avg}=\frac{146.01}{62}=2.35 m/min \approx  0.0391m/s$

Answer 2

The total distance traveled by an object divided by the total time taken is the average speed. The average velocity during a 62-minute period of time is 0.0391 m/sec.

What is the average speed?

The total distance traveled by an object divided by the total time taken is the average speed.

The average speed of an object indicates the pace at which it will traverse a distance. The unit of speed m/sec.

Cat firstly makes a displacement of 118 m towards north after that she travels 86 m towards west. Displacement is the shortest distance that's why the displacement is obtained by finding the resultant of the two displacements.

$\rm{Total \;displacement = \sqrt{(118)^{2}+(86)^{2}$

$\rm{Total \;displacement = \sqrt{21,230 }$

$\rm{Total \;displacement = 146.01 m}$

The vector form of the displacement is given by ;

$\vec{r} = -86 \vec{i}\;+\;118\vec{j}$

$\theta$ angle made by a vector with Y-axis

$\rm tan\theta$= $\frac{118}{86}$

$\rm tan\theta= 1.372$

$\theta = 53.91^0$ wrt to y-axis

$\theta\; = 180 - 53.91^0= 126.09 ^0$ wrt to the x-axis

$\rm V_ {avg} = \frac{\rm Total \;displacement }{\rm Total\;timer \;taken}$

$\rm V_ {avg} = \frac{\rm146.01 }{\rm 62}$

$\rm V_ {avg} = 2.35 m/min$

$\rm V_ {avg} = 0.0391 m/sec$

Hence the average velocity during a 62-minute period of time is 0.0391 m/sec.

To learn more about the average velocity refer to the link;

https://brainly.com/question/862972