Physics, asked by devansh4099, 5 months ago

a object moves on a circular path such that the distance covered is given by S=(0.5t^2 + 2t)m here t is time in seconds. if the radius of circle is 4m then the initial acceleration of object is​

Answers

Answered by nirman95
17

Given:

A object moves on a circular path such that the distance covered is given by S=(0.5t^2 + 2t)m here t is time in seconds. The radius of circle is 4m.

To find:

Initial acceleration of the object ?

Calculation:

Let tangential acceleration be a_(t):

 \therefore \: s = 0.5 {t}^{2}  + 2t

 \implies  \: v =  \dfrac{ds}{dt}

 \implies  \: v =  \dfrac{d(0.5 {t}^{2} + 2t) }{dt}

 \implies  \: v =  t + 2

 \implies \: a_{t} =  \dfrac{dv}{dt}

 \implies \: a_{t} =  \dfrac{d(t + 2)}{dt}

 \implies \: a_{t} = 1 \: m {s}^{ - 2}

So, tangential acceleration initially is 1 m/.

Let centripetal acceleration be a_(c):

 \therefore \: a_{c}  =  \dfrac{ {v}^{2} }{r}

 \implies \: a_{c}  =  \dfrac{ {(t + 2)}^{2} }{r}

Putting t = 0 secs for initial value:

 \implies \: a_{c}  =  \dfrac{ {(0 + 2)}^{2} }{4}

 \implies \: a_{c}  =  \dfrac{ 4}{4}

 \implies \: a_{c}  =  1 \: m {s}^{ - 2}

So, net acceleration:

 \therefore \: a_{net} =  \sqrt{ {(a_{t})}^{2}  +  {(a_{c})}^{2} }

 \implies \: a_{net} =  \sqrt{ {1}^{2}  +  {1}^{2} }

 \implies \: a_{net} =  \sqrt{2 }  \: m {s}^{ - 2}

So, net acceleration is 2 m/.

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