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

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

Smart Achievers · Accepted Answer

(3)

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

$a = \sqrt{a_{r}^{2} + a_{t}^{2}} = \sqrt{{(2.5)}^{2} + 2^{2}} = \sqrt{10.25} = 3.2 {m s}^{- 2}$

Solution

Q.
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

1 5 $m s^{- 2}$

2 2 $m s^{- 2}$

3 3.2 $m s^{- 2}$

4 4.3 $m s^{- 2}$

Ans.
(3)

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

$a = \sqrt{a_{r}^{2} + a_{t}^{2}} = \sqrt{{(2.5)}^{2} + 2^{2}} = \sqrt{10.25} = 3.2 {m s}^{- 2}$

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Solution

Q. 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

1 5 m s-2

2 2 m s-2

3 3.2 m s-2

4 4.3 m s-2

Ans. (3) Here, r=10 m, v=5 m s-1, at=2 m s-2 ar=v2r=5×510=2.5 m s-2 The net acceleration is a=ar2+at2=(2.5)2+22=10.25=3.2 m s-2