Recent content by mrkrgn

very nice! good answer

the problem was i was dividing by 0 when i differentiated with respect to x

ok i went about it a completely different way. this time rearranging it to x= y^2/4a now i will differentiate with respect to y instead of x. so dx/dy = 2y/4a → y/2a d^2x/dy^2 = 1/2a therefore curvature p = ((1+(y/2a)^2)3/2)/(1/2a) at the vertex which is (0,0)the curvature is 1*1/2a...

yep but i don't know how this would help me to find the max acceleration or how to go about it

yeah i did that and got it down to (1+a/x)^(9/4) = 0 which i assume is wrong

v= c/n c is the speed of light v is the speed in the new medium

do the collision between a and b first. total momentum before the impact equals the total momentum after, and the relative velocity after impact equals (-e)time relative velocity before impact. call initial velocities u1 and u2 and after collision v1 and v2 so, m1u1+ m2u2 = m1v1 + m2v2 and...

is there friction? as in is it actually rolling or sliding?

Homework Statement Hello, So i was just doing a few past papers for an upcoming exam and i got stuck on this one. i have got to the stage where i have 1/p all in term of x and y, but i can't see how you would go about getting the maximum acceleration just from x and y values...