derive the equation for postion-velocity relation
Answers
Explanation:
The first equation of motion relates velocity to time. We essentially derived it from this derivative… The second equation of motion relates position to time.
...
calculus derivations.
v = v0 + at [1]
+
s = s0 + v0t + ½at2 [2]
=
v2 = v02 + 2a(s − s0) [3]
v2 = v02 + 2a∆s [3]
method 2
The harder way to derive this equation is to start with the second equation of motion in this form…
∆s = v0t + ½at2 [2]
and solve it for time. This is not an easy job since the equation is quadratic. Rearrange terms like this…
½at2 + v0t − ∆s = 0
and compare it to the general form for a quadratic.
ax2 + bx + c = 0
The solutions to this are given by the famous equation…
x = −b ± √(b2 − 4ac)
2a
Replace the symbols in the general equation with the equivalent symbols from our rearranged second equation of motion…
t = −v0 ± √[v02 − 4(½a)(∆s)]
2(½a)
clean it up a bit…
t = −v0 ± √(v02 − 2a∆s)
a
and then substitute it back into the first equation of motion.
v = v0 + at [1]
v = v0 + a ⎛
⎜
⎝ −v0 ± √(v02 − 2a∆s) ⎠
a
Stuff cancels and we get this…
v = ±√(v02 + 2a∆s)
Square both sides and we're done.
v2 = v02 + 2a∆s [3]
Answer:
v = v0 + at [1]
+
s = s0 + v0t + ½at2 [2]
=
v2 = v02 + 2a(s − s0) [3]
Explanation:
The first equation of motion relates velocity to time. We essentially derived it from this derivative… The second equation of motion relates position to time.