Question

a position X=5t^3-3t^2 then the acceleration of the particle when velocity becomes zero is​

Answer 1

x = 5t³-3t²

we know,

instantaneous velocity v = dX/dt = 15t²-6t

A/Q

15t²-6t = 0

15t² = 6t

t = 6/15s

and

instantaneous acceleration a = dv/dt = 30t -6

Instantaneous acceleration is the acceleration when velocity approaches to zero.

acceleration when velocity becomes zero = 30×6/15 -6 = 6m/s²

Cheers!

Answer 2

Given:

Position of particle,x= 5t³-3t²

To Find:

Acceleration of the particle when velocity becomes zero

Solution:

We know that,

• Velocity of a body 'v' is given by

v = dx/dt

• Acceleration of a body 'a' is given by

a = dv/dt

where,

x is the displacement of particle at time t

━━━━━━━━━━━━━━━━━━━━━

Let the velocity of given particle be v and acceleration of given particle be a

So,

v= dx/dt

v= d(5t³-3t²)/dt

v= 5(3t²)-3(2t)

v= 15t²-6t

At v=0

15t²-6t= 0

3(5t²-2t)= 0

5t²=2t

5t= 2

t= 2/5 sec

So, at t= 2/5 sec, velocity of particle is zero

Now,

a= dv/dt

a= d(15t²-6t)/dt

a= 15(2t)-6(1)

a= 30t-6

At t= 2/5 sec

a= 30(2/5) -6

a= 12-6

a= 6m/s²

Hence, acceleration of the particle when velocity becomes zero is​ 6 m/s².