Question

Acceleration of a particle starting from rest is a = (6t) m/s-, then its velocity at t = 1 sec is

Answer 1

SOLUTION

GIVEN

Acceleration of a particle starting from rest is

$\sf{a = (6t) \: \: m {s}^{ - 2} }$

TO DETERMINE

The velocity at t = 1 sec

EVALUATION

Here it is given that the Acceleration of a particle starting from rest is

$\sf{a = (6t) \: \: m {s}^{ - 2} }$

Now

$\sf{a = (6t) \: }$

$\displaystyle \sf{ \implies \: \frac{dv}{dt} = 6t}$

$\displaystyle \sf{ \implies \: dv = 6t \: \: dt}$

On integration we get

$\displaystyle \sf{ \int dv = \int 6t \: \: dt}$

$\displaystyle \sf{ \implies v =6 \times \frac{ {t}^{2} }{2} + c }$

$\displaystyle \sf{ \implies v =3 {t}^{2} + c } \: \: - - - - (1)$

Now at t = 0 we have v = 0

Which gives

$\displaystyle \sf{ \implies 0 =3 \times {0}^{2} + c }$

$\displaystyle \sf{ \implies c = 0}$

So Equation 1 becomes

$\displaystyle \sf{ v = 3 {t}^{2} }$

When t = 1 we have

$\displaystyle \sf{ v = 3 \times {1}^{2} = 3 }$

Hence the required velocity at t = 1 sec

$\sf{ = 3 \: \: m {s}^{ - 1} }$

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