The velocity of a particle moving along the x-axis is v(t) = t^2 + 2t + 1, with t measured in minutes and v(t) measured in feet per minute. To the nearest foot find the total distance traveled by the particle from t = 0 to t = 2 minutes.
I took the definite integral of t^2+2t+1 from [0,2] and got 9. Is this correct? I'm not sure if I'm doing it correctly
not so.
∫[0,2] t^2+2t+1 dt
= t^3/3 + t^2 + t [0,2]
= (8/3 + 4 + 2)-0
= 26/3
Yes I got 26/3 as the exact answer but in terms of it being to the nearest foot it would be 9 right?
I'd say so, yes.
To find the total distance traveled by the particle from t = 0 to t = 2 minutes, you need to integrate the absolute value of the velocity function over the given interval.
First, let's find the absolute value of the velocity function v(t) = t^2 + 2t + 1 by simply removing the negative sign (if any):
|v(t)| = |t^2 + 2t + 1|
Next, we need to integrate the absolute value of the velocity function from t = 0 to t = 2. Integrating the absolute value of a function requires splitting the interval into smaller intervals where the function changes its sign. In this case, the function v(t) = t^2 + 2t + 1 never changes sign over the given interval [0, 2], so we can simply integrate the function as it is.
∫[0,2] |v(t)| dt = ∫[0,2] (t^2 + 2t + 1) dt
Evaluating the integral:
= [ (t^3)/3 + t^2 + t ] from 0 to 2
= (2^3)/3 + (2^2) + 2 - (0^3)/3 - (0^2) - 0
= (8/3) + 4 + 2
= 8/3 + 12/3 + 6/3
= 26/3
Converting the answer to the nearest foot:
26/3 feet is approximately equal to 8.67 feet.
Therefore, the total distance traveled by the particle from t = 0 to t = 2 minutes is approximately 8.67 feet.