Question 173: Motion in 2D & Vectors

Question

The positions vector of a particle $$\overrightarrow R $$ as a function of time is given by $$\overrightarrow R $$ = 4sin(2$$\pi $$t)$$\widehat i$$ + 4cos(2$$\pi $$t)$$\widehat j$$. Where R is in meters, t is in seconds and $$\widehat i$$ and $$\widehat j$$ denote unit vectors along x-and y-directions, respectively. Which one of the following statements is wrong for the motion of particle?

Accepted Answer

Here, $$\overrightarrow R  = 4\sin \left( {2\pi t} ight)\widehat i + 4\cos \left( {2\pi t} ight)\widehat j$$
The velocity of the particle is
$$\overrightarrow v  = {{d\overrightarrow R } \over {dt}} = {d \over {dt}}\left[ {4\sin \left( {2\pi t} ight)\widehat i + 4\cos \left( {2\pi t} ight)\widehat j} ight]$$
$$ = 8\pi cos\left( {2\pi t} ight)\widehat i - 8\pi sin\left( {2\pi t} ight)\widehat j$$
Its magnitude is
$$\left| {\overrightarrow v } ight| = \sqrt {{{\left( {8\pi \cos \left( {2\pi t} ight)} ight)}^2} + {{\left( { - 8\pi \sin \left( {2\pi t} ight)} ight)}^2}} $$
$$ = \sqrt {64{\pi ^2}{{\cos }^2}\left( {2\pi t} ight) + 64{\pi ^2}{{\sin }^2}\left( {2\pi t} ight)} $$
$$ = \sqrt {64{\pi ^2}\left[ {{{\cos }^2}\left( {2\pi t} ight) + {{\sin }^2}\left( {2\pi t} ight)} ight]} $$
$$ = \sqrt {64{\pi ^2}} $$     (as sin
2
$$	heta $$ + cos
2
$$	heta $$ = 1)
= 8$$\pi $$ m/s

Answer

Magnitude of the velocity  of particle is 8 meter/second.

Answer

Path of the particle is a circle of radius 4 meter.

Answer

Acceleration vector is along $$-$$$$\overrightarrow R $$.

Answer

Magnitude of acceleration vector is $${{{v^2}} \over R}$$ where v is the velocity of particle.

Motion in 2D & Vectors

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