Answer to Question #142165 in Classical Mechanics for Sridhar

Question

A particle of mass m=1Kg moves in the x -y plane.The force on it at time t is
F(t)= [2sin(αt)i+3cos(αt)j]N
where α =1s^(-1) At time t=0 the particle is at rest at the origin. Calculate the magnitude of its position vector r (in m ) and velocity vector v(in m/s) at
time t=(π/2)s
Ans r=[(π-2)²+9]^(1/2)
V=√13

Answer 1

Solution

Using newton equation

$m\frac{dV}{dt}=$  [2sin(αt)i+3cos(αt)j]N

Integrate above equation with respect to T

$V= \frac{[-2cos(αt)i+3sin(αt)j]}{\alpha m}+c$ .,,,,,,,,eq1

At initially t=0 V=0

We get c= -2i

So velcity become

$V= \frac{[-2cos(αt)i+3sin(αt)j]}{\alpha m}-2i$

Using $t=\pi/2$ and $\alpha=1$

$V=-2i+3j$

Magnitude of this velocity is

$V=\sqrt{13} m/s$

Again integrate equation 1 we get

Position vector

$r= \frac{[-2sin(αt)i+3cos(αt)j]}{\alpha^2m}-2it+3j$

Putting all given value

Position vector become as

$r=[(π-2)²+9]^{\frac{1}{2}}m$