deduse the equations of motion for constant acceleration using method of Calculus
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Calculus is an advanced math topic, but it makes deriving two of the three equations of motion much simpler. By definition, acceleration is the first derivative of velocity with respect to time. Take the operation in that definition and reverse it. Instead of differentiating velocity to find acceleration, integrate acceleration to find velocity. This gives us the velocity-time equation. If we assume acceleration is constant, we get the so-called first equation of motion [1].
a = dv dtdv = a dt vt⌠
⌡dv = ⌠
⌡a dt v00v − v0 = at v = v0 + at [1]
Again by definition, velocity is the first derivative of position with respect to time. Reverse the operation in the definition. Instead of differentiating position to find velocity, integrate velocity to find position. This gives us the position-time equation for constant acceleration, also known as the second equation of motion [2].
v = dsdtds = v dt = (v0 + at) dt st⌠
⌡ds = ⌠
⌡(v0 + at) dt s00s − s0 = v0t + ½at2 s = s0 + v0t + ½at2 [2]
Unlike the first and second equations of motion, there is no obvious way to derive the third equation of motion (the one that relates velocity to position) using calculus. We can't just reverse engineer the definitions. We need to play a rather sophisticated trick.
The first equation of motion relates velocity to time. We essentially derived it from this derivative…
dv = adt
The second equation of motion relates position to time. It came from this derivative…
d2s = adt2
The third equation of motion relates velocity to position. By logical extension, it should come from a derivative that looks like this…
dv = ?ds
But what does this equal? Well nothing by definition, but like all quantities it does equal itself. It also equals itself multiplied by 1. We'll use a special version of 1, dt/dt, and then do a bit of special algebra — algebra with infinitesimals. Look what happens when we do this. We get a derivative equal to acceleration and another equal to the inverse of velocity.
dv = dv · dt = dv · dt = a 1dsdsdtdtdsv
Next step, separation of variables. Get things that are similar together and integrate them. Here's what we get when acceleration is constant…
dv = a 1 dsv v dv = a ds vs⌠
⌡v dv = ⌠
⌡a ds v0s0½(v2 − v02) = a(s − s0) v2 = v02 + 2a(s − s0)
Certainly a clever solution, and it wasn't all that more difficult than the first two derivations. However, it really only worked because acceleration was constant — constant in time and constant in space. If acceleration varied in any way, this method would be uncomfortably difficult. We'd be back to using algebra just to save our sanity. Not that there's anything wrong with that. Algebra works and sanity is worth saving.
v = v0 + at [1] + s = s0 + v0t + ½at2 [2] = v2 = v02 + 2a(s − s0) [3]
a = dv dtdv = a dt vt⌠
⌡dv = ⌠
⌡a dt v00v − v0 = at v = v0 + at [1]
Again by definition, velocity is the first derivative of position with respect to time. Reverse the operation in the definition. Instead of differentiating position to find velocity, integrate velocity to find position. This gives us the position-time equation for constant acceleration, also known as the second equation of motion [2].
v = dsdtds = v dt = (v0 + at) dt st⌠
⌡ds = ⌠
⌡(v0 + at) dt s00s − s0 = v0t + ½at2 s = s0 + v0t + ½at2 [2]
Unlike the first and second equations of motion, there is no obvious way to derive the third equation of motion (the one that relates velocity to position) using calculus. We can't just reverse engineer the definitions. We need to play a rather sophisticated trick.
The first equation of motion relates velocity to time. We essentially derived it from this derivative…
dv = adt
The second equation of motion relates position to time. It came from this derivative…
d2s = adt2
The third equation of motion relates velocity to position. By logical extension, it should come from a derivative that looks like this…
dv = ?ds
But what does this equal? Well nothing by definition, but like all quantities it does equal itself. It also equals itself multiplied by 1. We'll use a special version of 1, dt/dt, and then do a bit of special algebra — algebra with infinitesimals. Look what happens when we do this. We get a derivative equal to acceleration and another equal to the inverse of velocity.
dv = dv · dt = dv · dt = a 1dsdsdtdtdsv
Next step, separation of variables. Get things that are similar together and integrate them. Here's what we get when acceleration is constant…
dv = a 1 dsv v dv = a ds vs⌠
⌡v dv = ⌠
⌡a ds v0s0½(v2 − v02) = a(s − s0) v2 = v02 + 2a(s − s0)
Certainly a clever solution, and it wasn't all that more difficult than the first two derivations. However, it really only worked because acceleration was constant — constant in time and constant in space. If acceleration varied in any way, this method would be uncomfortably difficult. We'd be back to using algebra just to save our sanity. Not that there's anything wrong with that. Algebra works and sanity is worth saving.
v = v0 + at [1] + s = s0 + v0t + ½at2 [2] = v2 = v02 + 2a(s − s0) [3]
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