A particle moves in a straight line with velocity 9−12𝑡+3𝑡^2 ft/s. Find the total total distance traveled over the time interval [1,6]
integral from t = 1 to t = 6
9 t - 6 t^2 + t^3 at t = 6 - at t = 1
at 6
9*6 - 6^3 + 6^3 = 54 (convenient)
at 1
9 - 6 + 1 = 4
54 - 4 = 50 feet
To find the total distance traveled over the time interval [1,6], we need to calculate the displacement for each time interval and sum them up.
First, we need to find the displacement function by integrating the velocity function over time.
Given velocity function: v(t) = 9 - 12t + 3t^2
Integrating the velocity function will give us the displacement function:
s(t) = ∫(9 - 12t + 3t^2) dt
To integrate this function, we use the power rule of integration:
∫t^n dt = (t^(n+1))/(n+1) + C
Applying the power rule to each term in the velocity function, we have:
s(t) = (9t - 6t^2 + t^3)/3 + C
Since we are given a time interval [1,6], we can find the displacement at t = 6 and t = 1 and subtract them.
Displacement at t = 6:
s(6) = (9*6 - 6*6^2 + 6^3)/3 + C = 18 + C
Displacement at t = 1:
s(1) = (9 - 6 + 1)/3 + C = 4 + C
Now, to find the total distance traveled, we need to consider the absolute value of the displacement, as the particle could have moved in either direction.
Total distance traveled = |s(6) - s(1)| = |18 + C - (4 + C)| = |14| = 14 feet
Therefore, the total distance traveled by the particle over the time interval [1,6] is 14 feet.