Physics - Circular Motion Question with Solution | TestHub

PhysicsCircular MotionMiscellaneousEasy2 minPYQ_2020

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A car is moving with speed $20 {m s}^{- 1}$ on a circular path of radius $100 m$ . Its speed is increasing at a rate of $3 {m s}^{- 2}$ . The magnitude of the acceleration of the car at that moment is

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Answer:

Solution:

Given $v = 20 {m s}^{- 1},$

$R = 100 m$

$\frac{d v}{d t} = 3 {m s}^{- 2}$

$a_{c} = \frac{v^{2}}{R} = \frac{{(20)}^{2}}{100} = 4 {m s}^{- 2}$

$a_{t} = \frac{d v}{d t} = 3 {m s}^{- 2}$

$a = \sqrt{a_{c}^{2} + a_{t}^{2}} = \sqrt{4^{2} + 3^{2}} = 5 {m s}^{- 2}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Circular MotionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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