Answer:
Therefore, the position of the particle when time = 1.33 seconds is approximately 15.253 meters.
Explanation:
To find the position of the particle at a given time, we need to integrate the velocity function with respect to time:
y = ∫(6t^3 - 2t^2 + 5) dt
y = 1.5t^4 - (2/3)t^3 + 5t + C (where C is the constant of integration)
Since we are given that the particle's velocity function is y = 6t^3 - 2t^2 + 5, we can determine C by evaluating the position function at t = 0:
y(0) = 1.5(0)^4 - (2/3)(0)^3 + 5(0) + C = C
Therefore, C = 0, and the position function is:
y = 1.5t^4 - (2/3)t^3 + 5t
To find the position of the particle when t = 1.33 seconds, we plug in that value of t:
y(1.33) = 1.5(1.33)^4 - (2/3)(1.33)^3 + 5(1.33)
y(1.33) ≈ 15.253 meters
Therefore, the position of the particle when time = 1.33 seconds is approximately 15.253 meters.