A 2.0 kg particle is moving along the x axis to the left with a velocity of v= -6.0 m/s. Suddenly, between times t =0 and t = 4.0 x a net force = 3t^2 – 12t is applied to the particle, where F is in N and t is in s. Calculate the velocity of the particle at t=4.0 s.

Is this formula right?
a= force/mass= (3t^2-12t)/4
v(4)=-12+a*t
now put t=4 into the equation

I get the acceleration as 0!!!

A 2.0 kg particle is moving along the x axis to the left with a velocity of v= -6.0 m/s. Suddenly, between times t =0 and t = 4.0 x a net force = 3t^2 – 12t is applied to the particle, where F is in N and t is in s. Calculate the velocity of the particle at t=4.0 s.

Is this formula right?

a= force/mass= (3t^2-12t)/2

if so, I get 0

How do I find the velocity at 4s?

To determine the velocity of the particle at t = 4.0 s, we need to calculate the acceleration at that time and then use it to find the final velocity.

The given formula for acceleration is correct:
a = force / mass = (3t^2 - 12t) / 4

Now, let's substitute t = 4.0 s into the equation to find the acceleration at t = 4.0 s:
a(4) = (3(4)^2 - 12(4)) / 4
= (3(16) - 48) / 4
= (48 - 48) / 4
= 0 / 4
= 0 m/s^2

So, you are correct in calculating the acceleration at t = 4.0 s as 0 m/s^2.

To determine the final velocity v(4), we can use the kinematic equation:
v(4) = v(0) + a * t

Given that the initial velocity v(0) is -6.0 m/s and the acceleration a(4) is 0 m/s^2, we substitute these values into the equation:
v(4) = -6.0 m/s + 0 m/s^2 * 4.0 s
= -6.0 m/s

Hence, the velocity of the particle at t = 4.0 s is -6.0 m/s.