The maximum velocity of the particle moving along the straight line is 36 units per time at t=3, while the minimum velocity is 0 units per time at t=0.
To find the maximum and minimum velocity of the particle moving along the straight line, we need to take the derivative of the position function and analyze its critical points.
1. Calculate the derivative of the position function, s(t):
- s'(t) = 4t³ - 12t² + 12t
2. To find the critical points, set the derivative equal to zero and solve for t:
- 4t³ - 12t² + 12t = 0
- Factor out a t: t(4t² - 12t + 12) = 0
- Apply the zero-product property: t = 0 or 4t² - 12t + 12 = 0
3. Solve the quadratic equation:
- Use the quadratic formula: t = (-b ± √(b² - 4ac)) / (2a)
- Plug in the values: t = (-(-12) ± √((-12)² - 4(4)(12))) / (2(4))
- Simplify: t = (12 ± √(144 - 192)) / 8
- Further simplify: t = (12 ± √(-48)) / 8
- Since the square root of a negative number is not real, there are no real solutions for t in this case.
4. Evaluate the velocity at the critical points and endpoints of the interval [0,3]:
- Calculate the velocity at t=0: s'(0) = 0
- Calculate the velocity at t=3: s'(3) = 4(3)³ - 12(3)² + 12(3)
- Simplify: s'(3) = 108 - 108 + 36 = 36
5. Determine the maximum and minimum velocities:
- Since there are no critical points within the interval [0,3], we only need to compare the velocity at the endpoints.
- The maximum velocity occurs at t=3 with a value of 36.
- The minimum velocity occurs at t=0 with a value of 0.
Therefore, the maximum velocity of the particle is 36 units per time, occurring at t=3. The minimum velocity is 0 units per time, occurring at t=0.
Note: Remember to include the proper units when discussing velocity.