Question

A particle is moving in a plane with a velocity given by

u=u◇icap (wa cos wt) j cap

where i and j are unit vectors along X- and Y-axes respectively. If the particle is at

the origin at 0, then its distance from the origin at time

t=3pi/2w, will be:

Answer 1

Answer:

Distance = $\boxed{\sqrt{(\frac{3\pi u_0}{2\omega})^2+a^2}}$

Explanation:

The velocity of the particle is given by

$\vec u=u_0\hat i+a\omega\cos\omega t\hat j$

at $t=0$ particle is at origin, hance $x=0$

We know that rate of change of displacement is velocity

Therefore,

$\boxed{\vec u=\frac{d\vec x}{dt}}$

$\implies d\vec x=\vec udt$

$\implies d\vec x=(u_0\hat i+a\omega\cos\omega t\hat j)dt$

$\implies d\vec x=u_0\hat idt+a\omega\cos\omega t\hat jdt$

$\implies \int_0^sd\vec x=(\int_0^{3\pi/2\omega}u_0dt)\hat i+(\int_0^{3\pi/2\omega}a\omega\cos\omega tdt)\hat j$

$\implies x\Bigr|_0^s=(u_0t\Bigr|_0^{3\pi/2\omega})\hat i+(a\omega\times\frac{1}{\omega}\sin\omega t\Bigr|_0^{3\pi/2\omega})\hat j$

$\implies \vec s=(u_0\frac{3\pi}{2\omega})\hat i+(a\sin\frac{3\pi}{2})\hat j$

$\implies \vec s=\frac{3\pi u_0}{2\omega}\hat i+a\hat j$

Therefore displacement

$=|\vec s|$

$=\sqrt{(\frac{3\pi u_0}{2\omega})^2+a^2}$

Hope this answer is helpful.

Answer 2

Answer:

nice question

Explanation:

You can see the above answer