Question 1189589: A particle moves in a straight line with velocity V(t)=root(3t-1) meters per second where t is time in seconds. At t=2, the particle's distance from the starding point was 8 meters in the positive direction. What is the particle's position at t=7 seconds?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem:
1. **Find the displacement function:**
The velocity function V(t) is the derivative of the position function s(t). To find the position function, we need to integrate the velocity function:
s(t) = ∫V(t) dt = ∫√(3t - 1) dt
Let u = 3t - 1, so du = 3 dt, and dt = du/3. Substituting:
s(t) = ∫√u * (du/3) = (1/3) ∫u^(1/2) du
s(t) = (1/3) * (2/3) * u^(3/2) + C = (2/9) * (3t - 1)^(3/2) + C
where C is the constant of integration.
2. **Solve for the constant of integration (C):**
We are given that at t = 2, the particle's position is 8 meters. Plug in these values to solve for C:
8 = (2/9) * (3*2 - 1)^(3/2) + C
8 = (2/9) * 5^(3/2) + C
8 = (2/9) * 5√5 + C
C = 8 - (10√5)/9
3. **Find the position at t = 7:**
Now that we have the complete position function, we can find the position at t = 7:
s(7) = (2/9) * (3*7 - 1)^(3/2) + 8 - (10√5)/9
s(7) = (2/9) * 20^(3/2) + 8 - (10√5)/9
s(7) = (2/9) * 20√20 + 8 - (10√5)/9
s(7) = (40√5)/9 + 8 - (10√5)/9
s(7) = (30√5)/9 + 8
s(7) = (10√5)/3 + 8
s(7) ≈ (10 * 2.236) / 3 + 8
s(7) ≈ 7.453 + 8
s(7) ≈ 15.453
Therefore, the particle's position at t = 7 seconds is approximately 15.45 meters.
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