I figured it out! There's an error in the text where they forgot to leave the ##\frac{1}{2\pi} ## in the numerator, and also I made an error in the summation earlier (counted zero twice!).
Ahh...I apologize, I forgot to add one more term to the equations:
##x_{A}(t) = x_{0,A} + v_{0,A}t + \frac{1}{2}a_{A}t^{2}##
##x_{B}(t) = x_{0,B} + v_{0,B}t + \frac{1}{2}a_{B}t^{2}##
We need to take into account the initial displacement between the two trains.
You should get that the trains...
Maybe you should start by trying to write the kinematic equations for both Train 1 and Train 2 given their velocities in terms of time. Then calculate the difference in distance as
##x_{A} - x_{B}##
The introductory physics equation you are looking for is:
##x(t) = v_{0}t +...
Your two equations are:
##\Delta X = (55 m/s) \times \cos(\theta)t##
##\Delta Y = (55 m/s) \times \sin(\theta)t - (4.9t^{2}) ##
This is a system of two equations with two variables, you solve through substitution or any other algebraic way of solving systems of equations. One such way is...
Homework Statement
I'm currently trying to follow a derivation done by Shankar in his "Basic Training in Mathematics" textbook. The derivation is on pages 343-344 and it is based on the solution to the two dimensional heat equation in polar coordinates, and I'm not sure how he gets from one...