Understanding Edexcel GCE January 2010 Mechanics M2 QP: Question 8C Explanation

[phospho]

http://www.scribd.com/doc/26846418/Edexcel-GCE-January-2010-Mechanics-M2-QP

http://www.edexcel.com/migrationdoc...CE January 2010 - MS/6678_01_msc_20100219.pdf

Question 8C:

The second link is the answers - I don't understand why they set x = 0 when they differentiated, could anyone explain please?

 

[TSny]

Generally, the derivative of a function is zero at a point where the function has a minimum.
[Edit:Sorry, please ignore my comment. I was looking at the first problem.]

 

[collinsmark]

http://www.scribd.com/doc/26846418/Edexcel-GCE-January-2010-Mechanics-M2-QP

http://www.edexcel.com/migrationdoc...CE January 2010 - MS/6678_01_msc_20100219.pdf

Question 8C:

The second link is the answers - I don't understand why they set x = 0 when they differentiated, could anyone explain please?

The solution is not differentiating nor "sets" x = 0 at that part of the solution.

What the solution is trying to say is:

"At whatever time it is when x happens to be zero, the y-component of velocity is c times greater than the x-component of velocity."

Recall that in the original problem statement, the initial velocity was given to you as u(i + cj). So the slope ("rise over run", also called dy/dx) is c/1 = c.

So at the very beginning, when the particle is launched, x = 0. (This is implied by the problem statement saying, "Relative to O, the position vector of a point on the path of P is (xi + yj) m.)

So still in other words, the statement is saying, "At the time the particle was launched (i.e. x = 0), the velocity slope (dy/dt over dx/dt) is c."