An object moves in such a manner that its velocity at time t is given by
v(t) = t3 - 4t2 + 3t
At t = 0, it starts at position x = 0. During the interval [0, 4] what is the net distance moved?
To find the net distance moved by the object during the interval [0, 4], we first need to find the displacement of the object over that interval. The displacement is the change in position and can be found by integrating the velocity function v(t) with respect to time from t = 0 to t = 4.
The displacement, Δx, is given by the definite integral:
Δx = ∫[0, 4] (t3 - 4t2 + 3t) dt
To solve this integral, we need to find the antiderivative (or the indefinite integral) of the velocity function.
The antiderivative of v(t) is given by:
∫ (t3 - 4t2 + 3t) dt = ¼ * t^4 - 4/3 * t^3 + 3/2 * t^2 + C
where C is the constant of integration.
Now, we can find the displacement over the interval [0, 4] by plugging in the bounds of integration into the antiderivative:
Δx = [¼ * t^4 - 4/3 * t^3 + 3/2 * t^2] [0, 4]
Substituting t = 4 into the antiderivative, we get:
Δx = (¼ * 4^4 - 4/3 * 4^3 + 3/2 * 4^2) - (¼ * 0^4 - 4/3 * 0^3 + 3/2 * 0^2)
Simplifying, we have:
Δx = (64 - 64 + 24) - (0) = 24
Therefore, the displacement of the object over the interval [0, 4] is 24 units.
Since the net distance is the absolute value of the displacement, the net distance moved by the object during the interval [0, 4] is |24| = 24 units.