Question

a particle starts from rest from origin and moves along a straight line according to the law a=mew cos t where a=acceleration and mew is constant .find it's displacement time t​

Answer 1

Please find the attached screenshot....

Answer 2

SOLUTION

GIVEN

A particle starts from rest from origin and moves along a straight line according to the law a = μ cos t where a = acceleration and μ is constant .

TO DETERMINE

The displacement at time t

EVALUATION

Here it is given that a particle starts from rest from origin and moves along a straight line according to the law a = μ cos t where a = acceleration and μ is constant

Thus we have

$\displaystyle\sf{a = \mu \cos t}$

$\displaystyle\sf{ \implies \: \frac{dv}{dt} = \mu \cos t}$

$\displaystyle\sf{ \implies \: dv = \mu \cos t \: \: dt}$

Integrating both sides we get

$\displaystyle\sf{ \implies \int dv = \int\mu \cos t \: \: dt}$

$\displaystyle\sf{ \implies v = \mu \sin t \: + c}$

Now at t = 0 we have v = 0

Thus we get c = 0

From above we get

$\displaystyle\sf{ \implies v = \mu \sin t \: }$

$\displaystyle\sf{ \implies \frac{ds}{dt} = \mu \sin t \: }$

$\displaystyle\sf{ \implies ds = \mu \sin t \: dt }$

Integrating we get

$\displaystyle\sf{ \implies \int \: ds = \int \: \mu \sin t \: dt }$

$\displaystyle\sf{ \implies s = - \: \mu \cos t \: + k }$

At t = 0 we have s = 0

Thus we get

$\displaystyle\sf{ \implies 0 = - \: \mu \: + k }$

$\displaystyle\sf{ \implies k = \: \mu \: }$

Thus we get

$\displaystyle\sf{ \implies s = - \: \mu \cos t \: + \mu }$

$\displaystyle\sf{ \implies s = \mu (1 - \cos t \: ) }$

Which is displacement at time t

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