Determining Velocity and Position Functions from Acceleration

Key Concepts

Integration: Used to find the velocity and position functions from the acceleration function.
Constants in Integration: Consideration of constants of integration when solving for functions.
Using Given Points: Use specific points to solve for constants in the functions.

Velocity function (V(t)) is determined by integrating the acceleration function.
Example acceleration function: 12t - 6.
- Use the power rule for integration.
- Antiderivative of t^1: ( t^2/2 ).
- Antiderivative of constant ( -6 ): ( -6t ).
- Add constant C after integration.
Velocity function: [ V(t) = 6t^2 - 6t + C ].
Use given point ( V(1) = 10 ) to solve for C.
- Substitute into equation: ( 6(1)^2 - 6(1) + C = 10 ).
- Solve: ( C = 10 ).
Final Velocity Function: [ V(t) = 6t^2 - 6t + 10 ].

Position function is found by integrating the velocity function.
Integrate: ( 6t^2 - 6t + 10 ).
- Antiderivative of ( 6t^2 ): ( 6t^3/3 ).
- Antiderivative of ( -6t ): ( 3t^2 ).
- Antiderivative of constant 10 with respect to t: ( 10t ).
Position function: [ X(t) = 2t^3 - 3t^2 + 10t + C ].
Use given point ( X(2) = 17 ) to solve for C.
- Substitute into equation: ( 2(2)^3 - 3(2)^2 + 10(2) + C = 17 ).
- Solve: ( C = -7 ).
Final Position Function: [ X(t) = 2t^3 - 3t^2 + 10t - 7 ].

Integration is key to transitioning from acceleration to velocity and then to position.
Use given points to find constants of integration.
These techniques are useful for solving physics and calculus problems involving motion.

If you want more examples or practice problems, check out the links in the description section of the video.