Recent content by joc

(1/2)vT = 2.pi.r -----(2) is an equation concerning distance travelled. the distance traveled is equal to the area under the speed-time graph, and is therefore (1/2)vT since acceleration is constant. the RHS is the circumference of the circle, i.e. the distance around the track. hope this...

this is an interesting question and i'd like to see someone shed some light on the solution as well.

actually integration in itself isn't particularly impressive (most people around me learn it at 16); in fact i just recalled that taking Further Math at 16-17 is actually normal and not exceptional, so i take my compliment back. no offence blue_chip :)

heh that's cool. then you'll be like, a match for some of the more accelerated people in the US :P enjoy yourself.

BLUE_CHIP: you're taking Further Maths at 16? as in the A-level subject? that's pretty impressive...

ahh yes, thanks doc.

darn, i got here too late :P i got D/L \leq \mu\sin\theta + \cos\theta but i didn't go any further. should've realized that applying the cosine rule was the next logical step. oof. krab, why do you use \pm\mu as the coefficient of B_x? doesn't friction always act in the same direction in...

i've been working on this problem...and it's basically a complicated trigonometry problem. you have to use several equations: 1) F_f\leq\mu\F_N 2) balance of vertical and horizontal forces 3) balance of torques 4) the sine rule (for both lengths and forces i think) i'm trying to do it...

draw the free body diagram - it's simply an object with 2 forces acting on it: tension and weight. letting T = tension in string, m = mass of object, w = weight of object = mg, v = speed of object, r = radius of rotation, A = angle that string makes with vertical, equating vertical...

you'd need both momentum and energy equations for this question. letting Mb = mass of bullet, M = total mass of can and bullet, Vb = velocity of bullet just before collision, V = velocity of bullet+can immediately after collision, h = height of rise, we can write 2 equations...

yes, the answer is d/2. actually i was looking for a simple method to solve the problem which didn't involve direct integration; but thanks, arcnets, for your help - it's useful to know how to solve general pursuit curve problems. :)

is the answer arccos [ [30 + sqrt(2)] / 32 ] = arccos [ 0.9817 ] = 79.0 degrees?

sorry for the misinformation. would 'thermodynamically favourable' be correct?

russ_watters put it nicely. acceleration is a physical result of force, but via that equation it is tied to velocity. sometimes in physics calculations, what directly causes a phenomenon is not as important as its relationship with various quantitative variables. anyway good for you that you've...

argh, no, the equation doesn't screw up in this case. it's exact for the model we're considering. eqn for drag force: Fd = -kv or Fd = -kv^2 (for high velocities) where k is a constant and v is the velocity. the negative sign indicates that the drag acts opposite to the direction of...