The velocity function, in feet per second, is given for a particle moving along a straight line.
v(t) = 3t − 2, 0 ≤ t ≤ 3
Find the displacement
15/2
Find the total distance that the particle travels over the given interval
Im not sure how to do this one
oops just backwards, since v < 0 until t = 2/3
To find the displacement of the particle, we need to integrate the velocity function over the given time interval.
The displacement is given by the formula:
Displacement = ∫ [v(t)] dt
∫ (3t - 2) dt
To solve this integral, we need to use the power rule of integration. The integral of t^n with respect to t is (1/n+1) t^(n+1).
Using this rule, we integrate each term separately:
∫ (3t - 2) dt
= (3/2) t^2 - 2t + C
Now, we evaluate the integral over the given interval (0 to 3):
Displacement = [(3/2) (3^2) - 2(3)] - [(3/2) (0^2) - 2(0)]
= (27/2 - 6) - (0 - 0)
= 27/2 - 6
= 15/2
Therefore, the displacement of the particle over the given interval is 15/2 feet.
Now, to find the total distance that the particle travels, we need to consider both the positive and negative values of velocity.
Since the velocity function (3t - 2) is positive for t > 2/3 and negative for t < 2/3, we need to split the integral into two parts and take the absolute value:
Total distance = ∫ |v(t)| dt
For t > 2/3,
∫ (3t - 2) dt = (3/2) t^2 - 2t + C
For t < 2/3,
∫ -(3t - 2) dt = -(3/2) t^2 + 2t + C'
Now, we evaluate each integral over the given interval (0 to 3):
Total distance = [(3/2) (3^2) - 2(3)] - [(3/2) (2/3)^2 - 2(2/3)] + [(3/2) (2/3)^2 - 2(2/3)]
= (27/2 - 6) - (2/2 - 4/3) + (2/2 - 4/3)
= 27/2 - 6 + 2/3 + 2/3
= 15/2 + 4/3 + 4/3
= 30/6 + 8/6 + 8/6
= 46/6
= 23/3
Therefore, the total distance that the particle travels over the given interval is 23/3 feet.
To find the displacement, we need to integrate the velocity function over the given time interval. The displacement is the accumulated change in position.
Given the velocity function v(t) = 3t - 2, we can integrate it to find the displacement. The integral of v(t) over the interval [0,3] will give us the change in position or displacement.
∫(0 to 3) (3t - 2) dt
To solve this integral, we need to apply the power rule of integration.
The integral of t^n with respect to t is (t^(n+1))/(n+1). Using this rule, we can simplify the integral as follows:
∫(0 to 3) (3t - 2) dt
= [(3/2)t^2 - 2t] from 0 to 3
Now, we substitute the upper and lower limits into the equation:
[(3/2)(3)^2 - 2(3)] - [(3/2)(0)^2 - 2(0)]
= [(3/2)(9) - 6] - [0 - 0]
= (27/2) - 6
= 15/2
Therefore, the displacement of the particle over the given interval is 15/2.
To find the total distance traveled, we need to consider the absolute value of the velocity function since distance cannot be negative. The total distance is the sum of all the distances covered over the given interval.
To find the distances, we need to evaluate the absolute value of the velocity function at every point in the interval [0,3], integrate those values, and add them up.
The distance function |v(t)| for the given velocity function v(t) = 3t - 2 is:
|v(t)| = |3t - 2| =
3t - 2 if 3t - 2 ≥ 0
-(3t - 2) if 3t - 2 < 0
For 0 ≤ t ≤ 3, the expression 3t - 2 is always positive. Therefore, |v(t)| = 3t - 2.
To find the total distance, we integrate the absolute value of the velocity function over the interval [0,3]:
∫(0 to 3) (3t - 2) dt
Using the same steps as before, we can find the integral:
∫(0 to 3) (3t - 2) dt
= [(3/2)t^2 - 2t] from 0 to 3
Applying the upper and lower limits:
[(3/2)(3)^2 - 2(3)] - [(3/2)(0)^2 - 2(0)]
= (27/2) - 6
= 15/2
Therefore, the total distance that the particle travels over the given interval is also 15/2.
displacement is ∫[0,3] v dt
but since v < 0 at t = 2/3,
total distance is ∫[9,2/3] v dt + ∫[2/3,3] -v dt