The velocity function is v(t)= t^2-5t+ 4 for a particle moving along a line. The position function s(t) is an antiderivative of the velocity function. Find the displacement, s(6) - s(-3), by the particle during the time interval [-3,6].
what is the derivative of 1/3 t^3-5/2 t^2+4t+ C ?
That's the problem. I have no idea. The question is exactly like i posted it here. I did the same thing but i realized i wasn't given a value of what it should equal to. I was hoping there a different way of solving this problem
Never mind i got it.
To find the displacement, s(6) - s(-3), of the particle during the time interval [-3,6], we need to evaluate the position function, s(t), at those two points and subtract the values.
The position function, s(t), is the antiderivative of the velocity function, v(t). To find s(t), we need to integrate the velocity function with respect to t.
Given: v(t) = t^2 - 5t + 4
To find s(t), we integrate v(t):
s(t) = ∫(t^2 - 5t + 4) dt
Let's integrate each term separately:
∫(t^2 - 5t + 4) dt = ∫t^2 dt - ∫5t dt + ∫4 dt
Integrating each term:
∫t^2 dt = (1/3)t^3 + C1 (where C1 is the constant of integration)
∫5t dt = (5/2)t^2 + C2 (where C2 is the constant of integration)
∫4 dt = 4t + C3 (where C3 is the constant of integration)
Now, we can find s(t) by adding up the integrals:
s(t) = (1/3)t^3 + C1 - (5/2)t^2 + C2 + 4t + C3
Now that we have the position function s(t), we can find s(6) and s(-3):
s(6) = (1/3)(6)^3 + C1 - (5/2)(6)^2 + C2 + 4(6) + C3
= 72 + C1 - 90 + C2 + 24 + C3
= C1 + C2 + C3 + 6
s(-3) = (1/3)(-3)^3 + C1 - (5/2)(-3)^2 + C2 + 4(-3) + C3
= -9 + C1 - 45/2 + C2 - 12 + C3
= C1 + C2 + C3 - 21/2
Finally, we can find the displacement, s(6) - s(-3), by subtracting the values:
s(6) - s(-3) = (C1 + C2 + C3 + 6) - (C1 + C2 + C3 - 21/2)
= 6 - (-21/2)
= 6 + 21/2
= 12 + 21
= 33
Therefore, the displacement of the particle during the time interval [-3,6] is 33 units.