Question:
A particle is moving on a circular path of radius $r$ with uniform speed $v$. What is the displacement of the particle after it has described an angle of $60^?$?
A particle is moving on a circular path of radius $r$ with uniform speed $v$. What is the displacement of the particle after it has described an angle of $60^?$?
Updated On: Jul 5, 2022
- $r\sqrt{2}$
- $r\sqrt{3}$
- $r$
- $2r$
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The Correct Option is C
Solution and Explanation
According to cosine formula
$cos\,60^{?}=\frac{r^{2}+r^{2}-x^{2}}{2r^{2}}$
$2r^{2}cos60^{?}=2r^{2}-x^{2}$
$x^{2}=2r^{2}-2r^{2}cos60^{?}$
$=2r^{2}\left[2sin^{2}\,30^{?}\right]=r^{2}$
$\therefore x=r$
Displacement $AB = x = r$
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Notes on Motion in a plane
Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration