Question

A particle undergoes two displacements. The first has a magnitude of 550 cm and makes an angle of 160 degrees with the positive x-axis. The resultant displacement has a magnitude of 725 cm and is directed at an angle of 75 degrees to the positive x-axis. Find the magnitude and direction of the second displacement.

Answer 1

magnitude of second vector is 871 and it makes angle of 36.019 degrees to the positive x-axis

Solution

In this question we have given

vector 1=
`(550cos160,550sin160)`

and resultant =
`(725cos75,725sin75)`

let second vector be V=
`(Vcos\theta,Vsin\theta)`

`(550cos160,550sin160)+(Vcos\theta,Vsin\theta)=(725cos75,725sin75)`

`(-516.830, 188.1)+(Vcos\theta, Vsin\theta) = (187.64, 700.29)`

`(Vcos\theta, Vsin\theta) = (187.64, 700.29)-(-516.830, 188.1)\\(Vcos\theta, Vsin\theta)=(704.47,512.19)`............(1)

on comparing x and y component of both sides of equation 1

`VCos\theta=704.47............(2)\\Vsin\theta=512.19`............(3)

Therefore,

divide equation 3 by equation (2)

`(Sin\theta)/(cos\theta) Sin\theta=(512.19)/(704.47) \\tan\theta=.727\\\theta=36.019`

put value of
`\theta` in equation (2)

`Vcos36.019=704.47\\V* .8088=704.47\\V=871.0`