How Do You Calculate the Constants A and C for a Rocket's Trajectory Equation?

[azn4lyf89]

A rocket is fired at an angle from the top of a tower of height 50.0m. Because of the designs of its engines, its position coordinates are of the form x(t)=A+Bt^2 and y(t)=C+Dt^3, where A, B, C, and D are constants. The acceleration of the rocket after 1.00s after firing is a= (4.00i+3.00j)m/s^2. Find the constants A, B, C, and D including their SI units.

I took the derivative of the position vectors twice to get the acceleration vector and plugged in 1.00s into t to find D=0.500m/s^3 and B=2.00m/s^2. I am stuck on where to go after that to get A and C.

 

[LowlyPion]

A rocket is fired at an angle from the top of a tower of height 50.0m. Because of the designs of its engines, its position coordinates are of the form x(t)=A+Bt^2 and y(t)=C+Dt^3, where A, B, C, and D are constants. The acceleration of the rocket after 1.00s after firing is a= (4.00i+3.00j)m/s^2. Find the constants A, B, C, and D including their SI units.

I took the derivative of the position vectors twice to get the acceleration vector and plugged in 1.00s into t to find D=0.500m/s^3 and B=2.00m/s^2. I am stuck on where to go after that to get A and C.

What is C. Wasn't it given to you? What is A then? Remember the equation describes position. Where was its initial position?

 

[baba89]

As a scientist, it is important to use the given information and equations to solve for the missing constants. In this case, we know that the acceleration of the rocket is given by the equation a= (4.00i+3.00j)m/s^2. This means that the acceleration in the x-direction is 4.00 m/s^2 and the acceleration in the y-direction is 3.00 m/s^2.

Using the equations x(t)=A+Bt^2 and y(t)=C+Dt^3, we can plug in the values for B and D that we found earlier and set the equations equal to the known values for the position of the rocket after 1.00s. This gives us the following equations:

x(1.00) = A + (2.00 m/s^2)(1.00 s)^2 = A + 2.00 m
y(1.00) = C + (0.500 m/s^3)(1.00 s)^3 = C + 0.500 m

Using the known height of the tower, we can also set the initial position of the rocket to be at (0, 50.0m). This gives us the following equations:

x(0) = A + (2.00 m/s^2)(0)^2 = A
y(0) = C + (0.500 m/s^3)(0)^3 = C

Now we have a system of equations with two unknowns (A and C) and four equations. We can solve this system by eliminating one variable, for example, by subtracting the equations for x(1.00) and x(0):

x(1.00) - x(0) = (A + 2.00 m) - A = 2.00 m

Similarly, we can subtract the equations for y(1.00) and y(0):

y(1.00) - y(0) = (C + 0.500 m) - C = 0.500 m

Now we have two equations with two unknowns, which we can solve using basic algebra. This gives us the values A=2.00m and C=0.500m.

Therefore, the constants for the position equations are A=2.00m, B=2.00m/s^2, C=0.