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 during the interval [0, 4], we need to calculate the total distance traveled by the object.
The net distance moved is the total distance traveled, regardless of the direction. It does not take into account whether the object changes direction during the interval.
To calculate the net distance, we can integrate the absolute value of the velocity function over the interval [0, 4].
First, let's find the integral of the velocity function:
∫(t^3 - 4t^2 + 3t) dt
To integrate this polynomial function, we can use the power rule of integration:
∫(t^n) dt = (t^(n+1))/(n+1) + C
Applying the power rule, we have:
∫(t^3) dt = (t^4)/4 + C1
∫(-4t^2) dt = (-4/3) * (t^3) + C2
∫(3t) dt = (3/2) * (t^2) + C3
Now, we can substitute the limits of integration [0, 4] into each term:
[(4^4)/4 + C1] - [(0^4)/4 + C1] + [(-4/3) * (4^3) + C2] - [(-4/3) * (0^3) + C2] + [(3/2) * (4^2) + C3] - [(3/2) * (0^2) + C3]
This simplifies to:
(256/4) + C1 - 0 + (-4/3) * 64 + C2 - 0 + (3/2) * 16 + C3 - 0
= 64 + C1 - (256/3) + C2 + 24 + C3
Since C1, C2, and C3 are arbitrary constants, their values will cancel out in the calculation of the net distance traveled.
Thus, the net distance moved during the interval [0, 4] is:
64 - (256/3) + 24
= 144/3 - 256/3 + 72/3
= -40/3
Therefore, the net distance moved during the interval [0, 4] is -40/3 units.
To find the net distance moved during the interval [0, 4], we first need to determine the displacement of the object for each segment of time within the given interval. We can then sum up the individual displacements to find the net distance.
To find the displacement of the object for a given time segment, we need to integrate the velocity function over that time segment. The displacement (Δx) at time t is given by the definite integral of the velocity function v(t) from the initial time (0 in this case) to the given time t.
So, for each time segment, we calculate the displacement as follows:
Δx = ∫[0, t] v(t) dt
Given the velocity function v(t) = t^3 - 4t^2 + 3t, we can calculate the displacement for different time segments within the interval [0, 4].
For t = 0 to t = 4:
Δx[0, 4] = ∫[0, 4] (t^3 - 4t^2 + 3t) dt
To find this integral, we need to break it down into separate terms and integrate each term individually.
∫ t^3 dt = (t^4)/4
∫ -4t^2 dt = - (4t^3)/3
∫ 3t dt = (3t^2)/2
So, now we can calculate the integral:
Δx[0, 4] = [(t^4)/4] - [(4t^3)/3] + [(3t^2)/2]
Evaluating this integral from t = 0 to t = 4, we get:
Δx[0, 4] = [(4^4)/4] - [(4^3)/3] + [(3(4^2))/2] - [(0^4)/4] + [(4^3)/3] - [(3(0^2))/2]
Simplifying further:
Δx[0, 4] = 64 - 64/3 + 24 - 0 + 0 + 0 = 64/3 + 24
So, the net distance moved during the interval [0, 4] is:
Net distance = 64/3 + 24 = 88/3
Therefore, the net distance moved during the interval [0, 4] is 88/3 or approximately 29.33 units.