The velocity of a particle moving along the x-axis is v(t) = t2 – 2t, with t measured in minutes and v(t) measured in feet per minute. To the nearest foot find the total distance travelled by the particle from t = 0 to t = 3 minutes..
This is just
∫[0,3] t^2 - 2t dt
Its Zero
Well, well, well, we have a particle on the move! Let's see how far it traveled.
To find the total distance traveled by our little friend, we need to calculate the displacement for each interval and then sum 'em up.
To make things easier, let's first find the velocity at the endpoints, which are t=0 and t=3. Plug those values into the velocity function:
v(0) = (0)^2 - 2(0) = 0 feet per minute
v(3) = (3)^2 - 2(3) = 9 - 6 = 3 feet per minute
Now, the distance traveled on a given interval is just the absolute value of the displacement. So let's calculate the displacement for each interval:
From t=0 to t=3: displacement = v(3) - v(0) = 3 - 0 = 3 feet
Voila! The total distance traveled by our particle in 3 minutes is approximately 3 feet. Keep rollin' and strollin', little particle!
To find the total distance traveled by the particle from t = 0 to t = 3 minutes, we need to evaluate the definite integral of the absolute value of the velocity function over the given time interval.
The velocity function v(t) = t^2 - 2t gives the rate of change of position with respect to time. To find the distance traveled, we need to consider the absolute value of the velocity function since distance traveled is always positive.
The definite integral of the absolute value of v(t) over the interval [0, 3] represents the total distance traveled. The integral can be set up as follows:
∫[0,3] |v(t)| dt
= ∫[0,3] |t^2 - 2t| dt
Since the integrand has different expressions depending on the interval, we need to split the interval [0,3] into two separate intervals, where v(t) changes its behavior:
∫[0,3] |t^2 - 2t| dt = ∫[0,2] (2t - t^2) dt + ∫[2,3] (t^2 - 2t) dt
Evaluate each integral separately:
∫[0,2] (2t - t^2) dt = [t^2 - (t^3/3)] evaluated from t = 0 to t = 2
Plugging in the values t = 2 and t = 0, we get:
[2^2 - (2^3/3)] - [0^2 - (0^3/3)]
= (4 - 8/3) - (0 - 0)
= (12/3 - 8/3) = 4/3
∫[2,3] (t^2 - 2t) dt = [(t^3/3) - t^2] evaluated from t = 2 to t = 3
Plugging in the values t = 3 and t = 2, we get:
[(3^3/3) - 3^2] - [(2^3/3) - 2^2]
= (27/3 - 9) - (8/3 - 4)
= (18/3 - 9) - (4/3 - 4)
= 6 - (4/3)
= 18/3 - 4/3 = 14/3
To find the total distance traveled, we add the distances from each interval:
Total distance traveled = ∫[0,2] (2t - t^2) dt + ∫[2,3] (t^2 - 2t) dt
= 4/3 + 14/3
= 18/3
= 6 feet
Therefore, the total distance traveled by the particle from t = 0 to t = 3 minutes is approximately 6 feet.
V = t^2 - 2t = 3^2 - 2*3 = 3 Ft/min.
d = V*t = 3Ft./min. * 3min. = 9 Ft.