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
so, I took the integral of the equation so from 0 to 2 of t^2+2t+1 dt
I got 8.6666 or 8 2/3 so my answer would be 9 if that's rounded to the nearest foot.
Yes, looks good.
Your approach to finding the total distance traveled by the particle is correct. To find the total distance, you need to take the integral of the absolute value of the velocity function over the interval from t = 0 to t = 2.
Let's work through the steps:
1. Start with the velocity function: v(t) = t^2 + 2t + 1.
2. Take the integral of the absolute value of v(t) over the interval from 0 to 2:
∫[0 to 2] |t^2 + 2t + 1| dt.
3. Separate the integral into two parts since the function inside the absolute value can be negative. So we have:
∫[0 to 2] (t^2 + 2t + 1) dt - ∫[0 to 2] (t^2 + 2t + 1) dt.
4. Evaluate the integrals:
∫[0 to 2] (t^2 + 2t + 1) dt = [(1/3)t^3 + t^2 + t] [0 to 2] = (8/3 + 4 + 2) - (0 + 0 + 0) = 22/3.
5. Substitute the result back into the integral:
∫[0 to 2] |t^2 + 2t + 1| dt = 22/3 - 22/3 = 0.
Since the result of the integral is 0, it means that the total distance traveled by the particle is 0. However, this may seem counterintuitive, so let's examine the situation more closely.
The particle's velocity function, v(t), is a second-degree polynomial that opens upwards, and it is always positive. This means that the particle is moving in a single direction, never turning back or changing its direction. Therefore, the total distance traveled by the particle from t = 0 to t = 2 is simply the displacement from the initial position to the final position at t = 2, which is given by:
∫[0 to 2] (v(t)) dt = ∫[0 to 2] (t^2 + 2t + 1) dt = [(1/3)t^3 + t^2 + t] [0 to 2] = (8/3 + 4 + 2) - (0 + 0 + 0) = 22/3.
Rounded to the nearest foot, the total distance traveled by the particle is 7 feet.
To find the total distance traveled by the particle from t=0 to t=2 minutes, you correctly took the integral of the velocity equation:
∫[0 to 2] (t^2 + 2t + 1) dt
Integrating each term separately, we have:
∫[0 to 2] t^2 dt + ∫[0 to 2] 2t dt + ∫[0 to 2] 1 dt
Integrating term by term, we get:
(t^3/3) + (t^2) + (t) evaluated from 0 to 2,
(2^3/3 + 2^2 + 2) - (0^3/3 + 0^2 + 0)
Simplifying, we have:
(8/3 + 4 + 2) - (0)
(8/3 + 4 + 2)
(8/3 + 12/3 + 2)
(20/3 + 2)
26/3
To round this to the nearest foot, we divide 26 by 3, which is approximately 8.67 feet. Rounding to the nearest foot, the total distance traveled by the particle is 9 feet.