Final answer:
The velocity function of the rocket is v(t) = 5t² - 4t³ + 50 m/s. The rocket reaches its peak at t = 0.83 seconds. The position function of the rocket is x(t) = (5/3) t³ - (1/4) t⁴ + 50t + 15 meters.
Step-by-step explanation:
To find the velocity function of the rocket, we need to integrate the given acceleration function. The given acceleration function is a(t) = 10t - 12t². To find the velocity function, we integrate a(t) with respect to t:
- ∫a(t) dt = ∫(10t - 12t²) dt
- v(t) = 5t² - 4t³ + C
Next, we can find the specific value of C by substituting the velocity at t = 0 into the equation. We know that v(0) = 50 m/s, which gives us:
- v(0) = 5(0)² - 4(0)³ + C = 0 + C
- 50 = C
So the velocity function of the rocket is: v(t) = 5t² - 4t³ + 50 m/s.
To find when the rocket reaches its peak, we need to find when the velocity function reaches its maximum value. We can do this by finding the derivative of the velocity function and setting it equal to zero:
- v'(t) = 10t - 12t² = 0
- 10t - 12t² = 0
- t(10 - 12t) = 0
- t = 0 or t = 10/12 = 0.83 s
Since the time cannot be negative, the rocket reaches its peak at t = 0.83 seconds.
If the rocket was launched from a height of 15 meters, the equation for the position function can be found by integrating the velocity function with respect to t:
- ∫v(t) dt = ∫(5t² - 4t³ + 50) dt
- x(t) = (5/3) t³ - (1/4) t⁴ + 50t + D
Next, we can find the specific value of D by substituting the position at t = 0 into the equation. We know that x(0) = 15 meters, which gives us:
- x(0) = (5/3) (0)³ - (1/4) (0)⁴ + 50(0) + D = 0 + 0 + 0 + D
- 15 = D
So the position function of the rocket is: x(t) = (5/3) t³ - (1/4) t⁴ + 50t + 15 meters.