Recent content by WillParadigm

not really, I'm still lost on this problem... gah, so much work for just one problem... UGH.

ok, I somewhat understand the problem a little better, but I did something wrong... a=g=9.8 since acceleration is constant? because the cable just snaps, so it's just gravity, and a can be set equal to a=Force/mass and mass is equal to weight/g which in this case, then 700 lbs/9.8 m/s^2...

An engineer claims to have a new kind of "safety" air-brake for elevators. If the cable should snap, a patented velocity-sensor turns on the brakes once the elevator hits 40 mph. The brakes are powerful enough to stop a 700 lb elevator in 1/20 of a second. How much force do the brakes exert...

are you a physics professor/major? because you should be... =) thank you very much, that makes 100% sense now.

ok, I can understand that part for finding time, superman's final height g = 9.8 m/s² and upward velocity is negative, since the gravitational acceleration, which acts downward, is always positive. so when superman reaches his maximum height, his velocity should be zero. right? So, vf = 0...

ok reedited, so I know what I'm looking for, but the question is, how do I determine how to find them, what equation am I missing?

Kinematics Problem. How do I get T to solve for V(i). PLEASE HELP 9.Superman is supposed to leap tall buildings in a single bound. Suppose that he obeys the normal laws of physics in this feat. Estimate how fast he should be moving when he leaves the ground so that he just clears the top of...

but that y intercept equation, I don't see how it helps, it just gives me an equation for a line I already saw, but not even the whole line...

couldn't the final velocity be any point along the line in the above graph? see that's what's confusing me the most, that there aren't any concrete numbers that I can see to use I mean, for the first part of the graph, I can see yes, the equation should be y=1/2x+15 because the slope is...

so it's (10-15)/(120-0) to get acceleration of -1/24 and I just graph 15x + (1/2)(-1/24)x^2? all I got was a straight line

wait, now you lost me... how does that help me graph something in x intervals of Time t (s) and y intervals of Position x (m)

the book I have does have a drawing of a x versus t graph, but it derives it from already having acceleration... and used the equation x=Initial Velocity x Time + 1/2 x Acceleration x Time^2

the question was this exactly... "7 Chapter 2, Question 22,and plot an x vs t graph of the motion." so I mean, the directions aren't that helpful, plot an x vs t graph of the motion

that's all I can get, just that it curves up from 0 but then I know there are other dips and peaks, and a level off since the velocity goes to zero, but I don't know how to draw it

yeahh, some of us were confused too, and did what the instructions on the pic show, but we're supposed to graph an x versus t graph of that picture showing the velocity, so it looks somewhat like an exponential growth graph...