Question

A Body Of Mass 1Kg Is Acted Upon By Force F= 2Sin3Πt I + 3Cos 3Πtj Find Its Position At T=1Sec If At T=0 It Is At Origin

Answer 1

HELLO DEAR,

GIVEN:-

m = 1kg

F = 2sin3Πt î + 3cos3Πtj

we know:- F = ma

=> F = 1 * a

=> F = a

now,

dv/dt = a

=> dv/dt = 2sin3Πtî + 3cos3Πtj

$\bold{ \int \limits_{0} ^{1} {dv} = \int \limits_{0}^{1} \{2sin3\pi \: t \: i \: + \: 3cos3\pi \: t \: j \}.dt}$

=> $\bold{ \int \limits_{0}^{1} \frac{dx}{dt} = [\frac{2}{3\pi} ( - \cos3\pi \: t)i \: + \: \frac{3}{3\pi} ( \sin3\pi \: t)j]_{0}^{^1}}$

=> $\int \limits_{ 0 } ^{1} \frac{dx}{dt}$= 2/3π(-cos3π + cos0)î + 1/π(0 - 0)j

=> v = 2/3π(1 + 1)î + 1/π(0)j = dx/dt

=> x = 4/3π

and y = 0

I HOPE IT'S HELP YOU DEAR,

THANKS