Question

The position of a particle moving in a straight line is given by y = 3t^3 + t^2 + 5 (y in cm and t in second) then, find the acceleration of the particle at time t =2 second.

Answer 1

HEY BUDDY..!!!

HERE'S THE ANSWER..

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♠️ As we know , Velocity of a object is give by rate of change of displacement with respect to time , i.e

✔️ velocity = d ( y ) / d t

⏺️ Where ( y ) is displacement and ( t ) is time .


♠️ We know rate of change of velocity with respect to time is known as Acceleration , i.e

✔️ Acceleration = d ( v ) / d t

⏺️ Where ( v ) is velocity and ( t ) is time


▶️ Now in given question we have to double differentiate to find Acceleration from displacement , So here we go !!!

▶️ Formula for differentiation

✔️{ d ( x )^n / d x = n . ( x )^n - 1 } , n is constant

=> Displacement ( y ) = 3 ( t )^3 + ( t )^2 + 5

⏺️ By differentiating with respect to time ( t )

=> d ( y ) / d t = d ( 3 ( t )^3 + ( t )^2 + 5 ) / d t

=> velocity ( v ) = 9 ( t )^2 + 2 t

⏺️ Now to find Acceleration we'll differentiatiate velocity

=> d ( v ) / d t = d ( 9 ( t )^2 + 2 t ) / d t

=> Acceleration = 18 t + 2

▶️ Now Acceleration at 2 second, we'll put ( t = 2 )

=> Acceleration = 18 ( 2 ) + 2

=> Acceleration = 38 cm / s^2

♠️ In standard form

=> [ Acceleration = 0.38 m / s^2 ]



HOPE HELPED..

JAI HIND..

:-)

Answer 2

Explanation:

hope it helps you!!!!!!