. 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. Please show how to answer this problem. Thanks
To calculate the velocity of the particle at t=4.0 s, we need to find the time-dependent acceleration and then integrate it to obtain the change in velocity over the given time interval.
The given net force is given as F = 3t^2 – 12t N. Since F = ma, where m is the mass of the particle and a is the acceleration, we can equate them to get the time-dependent acceleration:
3t^2 – 12t = ma
The mass of the particle is given as 2.0 kg, so we can substitute this value in the equation:
3t^2 – 12t = 2a
Now, we can solve this differential equation to find the acceleration at any given time. We can use calculus to do so.
Taking the derivative of acceleration with respect to time gives us:
d²x/dt² = a = (6t - 12)
Integrating both sides of the equation with respect to time gives us the expression for velocity:
∫ d²x/dt² dt = ∫ (6t - 12) dt
Integrating the right side of the equation:
∫ (6t - 12) dt = 3t^2 - 12t + C
Where C is the constant of integration. We need to find the value of C using the given initial velocity of -6.0 m/s at t = 0.
Plugging in t = 0 and v = -6.0 into the expression:
3(0)^2 - 12(0) + C = -6.0
0 - 0 + C = -6.0
C = -6.0
Now we can find the velocity at t=4.0s by plugging t=4.0 into the expression:
v = 3(4.0)^2 - 12(4.0) - 6.0
v = 48.0 - 48.0 - 6.0
v = -6.0 m/s
Therefore, the velocity of the particle at t=4.0 s is -6.0 m/s.