Anyone amongst youse artist types what still remembers how to solve calculus problems?

Working on an accident reconstruction animation and the pointy-headed expert reconstructionist is not available over the holiday weekend.

I need to determine the distance a decelerating object will travel in 5 seconds if it starts at 30 mph (44ft/sec) and ends at 5mph (7.33ft/sec).

Sitting here on the horns of a dilemma, with a cryptic old calculus textbook on one side of the room, and a tape measure and car keys on the other side.

Nance, (a) Firstly, to find the deceleration, use : v = u + (at) --(1) where v is the velocity at any time t, u is the initial velocity (in this case 44). Substitute v=7.33 when t=5 into (1), we get : 7.33 = 44 + (5a) which, when re-arranging, gives : a = (1/5)(7.33 - 44) ... or a = -7.334 ft/sec^2 --(2) (b) To find the distance travelled, use : x = (ut) + (1/2)(at^2) --(3) Substituting u=44, t=5, a=-7.334 (from eq (2)) : x = (445) + (1/2)(-7.33455) ... or x = 128.325 ft. QED.

Calculus shouldn't be required. Simple Physics formulas should suffice. s = v(i)*t + (1/2)at^2 where: s is distance v(i) is initial velocity t is time interval a is acceleration Since you have final velocity [v(f)] and not a [acceleration], a is given by (v(f) - v(i)) / t. So, s = v(i)t + (1/2)(v(f) - v(i))*t :)

D'oh, I figured it out, but forgot the 1/2 coefficient. I always do that. :P Anyway, I can second Mathman's result there. I got 128.325 ft independently (once I remembered the 1/2 coefficient before the acceleration, that is.)

What an amazing place this is. Thanks to you kind, knowledgeable & fast folks! :-)

I can count in base eight if I use my toes ....