Question

$The \: acceleration \: 'a' \: of\: a \: \\ particle \: in \: m/ {s}^{2} \: is \: given \: by \\ a = 18 {t}^{2} + 36t + 120. \\ If \: the \: particle \: starts \: from \: \\ rest \: then \: what \: will \: be \: the \: \\ velocity \: at \: the \: end \: of \: 1s \: ?$

Answer 1

HEY BUDDY..!!


HERE'S THE ANSWER..

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♠️ We know time rate of change of change of velocity is know as ACCELERATION ( a ) , i.e

✔️ d ( v ) / d t = a

⏺️ Now to get velocity from Acceleration we have to integrate Acceleration with respect to time .


=> v = integration ( 18 t^2 + 36 t + 120 ) d t

=> v = 18 t^3 / 3 + 36 t^2 / 2 + 120 t

=> [ v = 6 t^3 + 18 t^2 + 120 t ]

▶️ Now from 1 second we'll put t = 1

=> v = 6 ( 1 )^3 + 18 ( 1 )^2 + 120 ( 1 )

=> v = 6 + 18 + 120

=> v = 144 m / s


HOPE HELPED..

JAI HIND..

:-)

Answer 2

Hey there !

Thanks for the question !

Solution:

We know that acceleration can be calculated as change in velocity per unit time.

In terms of calculus,

a = dv / dt

=> a.dt = dv

Now integrating on both sides we get,

=> ∫ a.dt = ∫ dv

=> ∫ ( 18t² + 36t + 120 ) dt = v

Applying Power rule we get,

=> ∫ 18 ( t²⁺¹ / 2 + 1 ) + 36 ( t¹⁺¹ / 1 + 1 ) + 120 ( t⁰⁺¹ / 0 + 1 ) . ∫ dt = v

=> 18 t³ / 3 + 36 t² / 2 + 120 t . 1 = v [ Since ∫ dt = 1 ]

=> 6t³ + 18t² + 120t = v

Now we need to find the velocity at t = 1 second

Therefore substituting t = 1 we get,

=> 6 ( 1 )³ + 18 ( 1 )² + 120 ( 1 ) = v

=> 6 + 18 + 120

=> 144 = v

Hence the velocity of particle 'a' in t = 1 second is 144 m/s.

Hope my answer helped !