Kinematics (physics with calculus)

Question

A rocket traveling upward has a height above the ground given by x(t)=At^2-Bt^3, where A and B are positive constants.

What is the maximum upward velocity the rocket will reach, in terms of A and B?

Answer 1

You have an expression for the position the object will reach versus time.

A dx/dt is an expression for the velocity.

 

dx/dt = v = 2At = 3Bt^2

 

To maximize this expression we take the derivative of the expression and set that equal to zero.

 

dv/dt = a = 2A -6Bt = 0;    t = 2A/6B =  A/(3B)  The velocity at this corresponding time can be found by substituting the time back into the velocity equation.

 

v(A/3B) = 2A (A/3B) -3*B (A/3B)^2 = 2A^2/3B -  3A^2/9B = A^2/(3B) 

 

We know that this time is a maximum because the second derivative of velocity is always negative.

 

da/dt = -6B.

Answer 2

To solve this problem we need to use the following principles: 

 

for any function f(x) the critical points occur where f'(x) = 0 or f'(x) does not exist.

if f''(x) < 0, then it is a local maximum, if f''(x) > 0, then it is a local minimum.

 

 

Position Equation: X(t) = At² - Bt³           dx/dt = 2At - 3Bt²

Velocity Equation: V(t) = 2At -3Bt²             d²x/dt² = 2A - 6Bt

Acceleration Equation: A(t) = 2A -6Bt

 

First we must find the critical points of the velocity equation. This will gives us the information we need to find the maximum and/or minimums.

 

The critical points occur where the derivatives of the graph equal zero.

Setting the acceleration equation to zero and solving for t:

 

2A - 6Bt = 0 -> t = A/3B.

 

At point t = A/3B, the acceleration of the rocket will equal zero.

To determine if it is a local maximum or minimum apply the principles above.

 

The derivation of the acceleration equation is: -6B, since B is a positive number -6B is negative.

Therefore, it is a local maximum. This is the highest velocity the rocket will reach. 

 

Plugging t = A/3B back into the velocity equation yields: 2A(A/3B) - 3B(A/3B)² = 2A²/3B - 3A²B/9B² = A²/3B

 

The rocket reaches maximum velocity A²/3B.