Question:
A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is ${5\, m \, s^{-1}}$ and the speed is increasing at a rate of ${2\, m \, s^{-2}}$. The magnitude of net acceleration at this instant is
A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is ${5\, m \, s^{-1}}$ and the speed is increasing at a rate of ${2\, m \, s^{-2}}$. The magnitude of net acceleration at this instant is
Updated On: Jun 7, 2022
- ${5\, m \, s^{-2}}$
- ${2\, m \, s^{-2}}$
- $3.2 \, ms^{-2}$
- $4.3 \, ms^{-2}$
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The Correct Option is C
Solution and Explanation
Here, $r = 10{m}, v = 5{m \, s^{-1}}, a_t = 2 {m \, s^{-2}},$
$a_r = \frac{v^2}{r} = \frac{5 \times 5}{10} = 2.5 {m \,s^{-2}}$
The net acceleration is
$ a = \sqrt{a^2_r + a^2_t} = \sqrt{(2.5)^2 +2^2}$
$= \sqrt{10.25} = 3.2 \, ms^{-2}$
$a_r = \frac{v^2}{r} = \frac{5 \times 5}{10} = 2.5 {m \,s^{-2}}$
The net acceleration is
$ a = \sqrt{a^2_r + a^2_t} = \sqrt{(2.5)^2 +2^2}$
$= \sqrt{10.25} = 3.2 \, ms^{-2}$
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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