Why do we use this special result... (1 Viewer)

Carl5 · May 14, 2011

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My book doesn't explain it. Not that it explained much in the first place hah.
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Pyrobooby · May 14, 2011

It lets you find the acceleration function.

hscishard · May 15, 2011

Because then we are able to make velocity-displacement and acceleration-displacement functions. This is helpful in situations where you want to find what the velocity/acceleration is at a certain distance from the origin.

Proof would be acceleration = dv/dt
=dv/dx * dx/dt [chainrule]
=dv/dx * v [dx/dt = velocity]
=dv/dx * d(v^2/2)/dv [v = d(0.5v^2)/dv]
=d(v^2/2)/dx = acceleration [the dv cancelled out]

schroot_92 · May 25, 2011

Simple differentiation and integration allows us to find acceleration, velocity and displacement in terms of time.. v = dx/dt and a = dv/dt.. By using this special result, we can find acceleration and velocity in terms of x(displacement)

Why do we use this special result... (1 Viewer)

Carl5

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