Assume v(t) = 6t - 18 on the interval [0, 6]. Find the total distance that the object traveled on [0, 6].
we just did one of these
x(t) = 3t^2 - 18t + c
Now just figure x(6)-x(0)
I got 0. That doesn't make sense to me...
x(0) = 3(0)^2 - 18(0) + C = 0+C
x(6) = 3(6)^2 - 18(6) + C = 108-108+C = 0+C
x(6)-x(0)=0
Your velocity is zero at t = 3 !!!!!
right in the middle of your range.
It reversed direction midstream
You calulated displacement (where it moved to which was nowhere)
instead of how far it went (distance moved)
do t = 0 to t = 3
then t = 3 to t = 6
add the ABSOLUTE VALUES
See Baggins, Bilbo, "There and Back Again".
So is the total distance 54?
yes
To find the total distance that the object traveled on the interval [0, 6], we can use the concept of distance traveled.
The distance traveled by an object can be determined by taking the absolute value of the velocity function and integrating it over the given interval.
In this case, the velocity function is v(t) = 6t - 18.
Step 1: Take the absolute value of the velocity function
To find the absolute value of 6t - 18, we need to determine the critical points where the function changes sign:
6t - 18 = 0
6t = 18
t = 3
So, the function changes sign at t = 3.
For t < 3, the velocity function is negative: v(t) = -(6t - 18) = -6t + 18.
For t > 3, the velocity function is positive: v(t) = 6t - 18
Step 2: Integrate the absolute value of the velocity function over the interval [0, 6]
We need to split the integration into two parts, from 0 to 3 and from 3 to 6:
For t < 3:
∫[0,3] |v(t)| dt = ∫[0,3] |-6t + 18| dt
For t > 3:
∫[3,6] |v(t)| dt = ∫[3,6] |6t - 18| dt
Step 3: Evaluate the integrals
For t < 3:
∫[0,3] |-6t + 18| dt = ∫[0,3] (6t - 18) dt = [3t^2 - 18t] from 0 to 3
= [(3(3)^2 - 18(3)) - (3(0)^2 - 18(0))]
= [9 - 54 - 0 + 0]
= -45
For t > 3:
∫[3,6] |6t - 18| dt = ∫[3,6] (6t - 18) dt = [3t^2 - 18t] from 3 to 6
= [(3(6)^2 - 18(6)) - (3(3)^2 - 18(3))]
= [108 - 108 - 27 + 54]
= 27
Step 4: Find the total distance traveled
To find the total distance traveled, we add the absolute values of the integrals for the two intervals together:
Total distance = |-(∫[0,3] |-6t + 18| dt)| + (∫[3,6] |6t - 18| dt)
= |-45| + 27
= 45 + 27
= 72
Therefore, the object traveled a total distance of 72 units on the interval [0, 6].