It is known that an object traveling in straight line with velocity function v(t) = t^2-t, where v is measured in meters per second. Find following.
The position function s(t) when we know the position 5 meters after 2 seconds.
I tried to work this out and got s(t) = (t^3)/3 -(t^2)/2 + 13/3.
Find the displacement and total distance traveled from [0,6].
Displacement I got 54 meters, but I'm not sure how to find total distance.
Can someone help me please
To find the position function s(t), you need to integrate the velocity function v(t) with respect to time t.
Start by integrating the velocity function:
∫(t^2 - t) dt
Using the power rule for integration, we obtain:
(t^3/3) - (t^2/2) + C
Since we know s(2) = 5 meters, we can use this information to find the constant C. Substituting t = 2 and s(t) = 5 into the position function, we have:
(2^3/3) - (2^2/2) + C = 5
Simplifying, we get:
8/3 - 2 + C = 5
8/3 - 6/3 + C = 5
2/3 + C = 5
C = 5 - 2/3
C = 13/3
So, the position function is:
s(t) = (t^3/3) - (t^2/2) + 13/3
Now, to find the displacement and total distance traveled from [0,6], we need to find the displacement first.
Displacement is the change in position from the initial time t = 0 to the final time t = 6. Therefore, we calculate s(6) - s(0):
s(6) - s(0) = [(6^3/3) - (6^2/2) + 13/3] - [(0^3/3) - (0^2/2) + 13/3]
= [216/3 - 36/2 + 13/3] - 0
= 72 - 18 + 13/3
= (36 + 13)/3
= 49/3
So, the displacement is 49/3 meters.
To find the total distance traveled, we need to consider both positive and negative displacements. To do this, we can take the absolute value of the displacement.
Total distance traveled = |49/3| = 49/3 meters.
Therefore, the displacement is 49/3 meters and the total distance traveled from [0,6] is also 49/3 meters.
To find the position function, you need to integrate the velocity function.
Given that the velocity function is v(t) = t^2 - t, you can integrate it to find the position function s(t):
∫ v(t) dt = ∫ (t^2 - t) dt
Using the power rule of integration, you can integrate both terms separately:
∫ t^2 dt - ∫ t dt
Integrating each term:
(t^3)/3 - (t^2)/2 + C
Where C is the constant of integration.
Now, you can find the position function s(t) when the position is 5 meters after 2 seconds by substituting t = 2 into the position function:
s(2) = (2^3)/3 - (2^2)/2 + C
Since the position is known to be 5 meters after 2 seconds, you can solve for the constant of integration C:
5 = (8/3) - 2 + C
5 = 8/3 - 6/3 + C
5 = 2/3 + C
C = 5 - 2/3
C = 15/3 - 2/3
C = 13/3
Therefore, the position function is:
s(t) = (t^3)/3 - (t^2)/2 + 13/3
To find the displacement from t = 0 to t = 6, you need to evaluate s(t) at those two points:
Displacement = s(6) - s(0)
= [(6^3)/3 - (6^2)/2 + 13/3] - [(0^3)/3 - (0^2)/2 + 13/3]
= [216/3 - 36/2 + 13/3] - [0/3 - 0/2 + 13/3]
= 72 - 18 + 13 - 0 - 0 + 13
= 80 meters
To find the total distance traveled from t = 0 to t = 6, you need to consider both the positive and negative portions of the position function.
You can observe that the velocity function v(t) = t^2 - t changes signs at t = 0 and t = 1.
So, you need to calculate the integral of the absolute value of the velocity function over the interval [0, 1] and the interval [1, 6].
∫ |v(t)| dt = ∫ |t^2 - t| dt
For the first interval, [0, 1], the function becomes:
∫ (t^2 - t) dt
Integrating each term separately:
(t^3)/3 - (t^2)/2
Evaluating the integral at the boundaries:
[(1^3)/3 - (1^2)/2] - [(0^3)/3 - (0^2)/2]
= [1/3 - 1/2] - [0/3 - 0/2]
= 1/3 - 1/2
= 2/6 - 3/6
= -1/6
For the second interval, [1, 6], the function becomes:
∫ (t^2 - t) dt
Integrating each term separately:
(t^3)/3 - (t^2)/2
Evaluating the integral at the boundaries:
[(6^3)/3 - (6^2)/2] - [(1^3)/3 - (1^2)/2]
= [216/3 - 36/2] - [1/3 - 1/2]
= 72 - 18 - 1/3 + 1/2
= 72 - 18 - 2/6 + 3/6
= 54 - 2/6
= 54 - 1/3
= 53 2/3
Adding the absolute values of the integrals for the two intervals:
|-1/6| + |53 2/3|
= 1/6 + 53 2/3
= 321/6 + 53 2/3
= 53 3/6 + 53 2/3
= 106/2 + 106/3
= (212 + 212)/6
= 424/6
= 70 2/3 meters
Therefore, the total distance traveled from t = 0 to t = 6 is 70 2/3 meters.