ShawnD1
02-28-2003, 03:49 AM
the formula for kinetic energy is 1/2 mv^2
the only thing i can't figure out is the 1/2
if you draw a graph of v as x and E as y, it's clearly a quadratic. since the mass is a constant in the graph, it can be ignored. the derivative for the graph would be the slope at any given point and if you calculate it out, it happens to be (1/2 x v) (hence the v^2)....but why?
how does this 1/2 somehow find its way into the equation?
the formula works through actual lab experiments, but why?
ok now for final velocity based on distance, the formula is Vf^2 = Vi^2 + 2ad
if we assume we are dropping the item, we ignore Vi. now if you graph this as d on the x and Vf on the y (not Vf^2), you get a quadratic; that is not the real equation though, it's just a referance for how the graph should look. now to correct the graph we put square root of d on the x and Vf on the y. now if we take the derivative of that, we get that the slope is 2. we have proved that the 2 is there... but why?
the only thing i can't figure out is the 1/2
if you draw a graph of v as x and E as y, it's clearly a quadratic. since the mass is a constant in the graph, it can be ignored. the derivative for the graph would be the slope at any given point and if you calculate it out, it happens to be (1/2 x v) (hence the v^2)....but why?
how does this 1/2 somehow find its way into the equation?
the formula works through actual lab experiments, but why?
ok now for final velocity based on distance, the formula is Vf^2 = Vi^2 + 2ad
if we assume we are dropping the item, we ignore Vi. now if you graph this as d on the x and Vf on the y (not Vf^2), you get a quadratic; that is not the real equation though, it's just a referance for how the graph should look. now to correct the graph we put square root of d on the x and Vf on the y. now if we take the derivative of that, we get that the slope is 2. we have proved that the 2 is there... but why?