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Physics Question on Motion in a plane

A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is 5 m s−1{5\, m \, s^{-1}} and the speed is increasing at a rate of 2 m s−2{2\, m \, s^{-2}}. The magnitude of net acceleration at this instant is

A

5 m s−2{5\, m \, s^{-2}}

B

2 m s−2{2\, m \, s^{-2}}

C

3.2 ms−23.2 \, ms^{-2}

D

4.3 ms−24.3 \, ms^{-2}

Answer

3.2 ms−23.2 \, ms^{-2}

Explanation

Solution

Here, r=10m,v=5m s−1,at=2m s−2,r = 10{m}, v = 5{m \, s^{-1}}, a_t = 2 {m \, s^{-2}},
ar=v2r=5×510=2.5m s−2a_r = \frac{v^2}{r} = \frac{5 \times 5}{10} = 2.5 {m \,s^{-2}}
The net acceleration is
a=ar2+at2=(2.5)2+22a = \sqrt{a^2_r + a^2_t} = \sqrt{(2.5)^2 +2^2}
=10.25=3.2 ms−2= \sqrt{10.25} = 3.2 \, ms^{-2}