write velocity time equation of
motion
Answers
Answer:
method 1
Combine the first two equations together in a manner that will eliminate time as a variable. The easiest way to do this is to start with the first equation of motion…
v = v0 + at [1]
solve it for time…
t = v − v0
a
and then substitute it into the second equation of motion…
s = s0 + v0t + ½at2 [2]
like this…
s = s0 + v0 ⎛
⎜
⎝ v − v0 ⎞
⎟
⎠ + ½a ⎛
⎜
⎝ v − v0 ⎞2
⎟
⎠
a a
s − s0 = vv0 − v02 + v2 − 2vv0 + v02
a 2a
2a(s − s0) = 2(vv0 − v02) + (v2 − 2vv0 + v02)
2a(s − s0) = v2 − v02
Make velocity squared the subject and we're done.
v2 = v02 + 2a(s − s0) [3]
This is the third equation of motion. Once again, the symbol s0 [ess nought] is the initial position and s is the position some time t later. If you prefer, you may write the equation using ∆s — the change in position, displacement, or distance as the situation merits.
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]
Explanation:
hope it helps you
Answer:
let's devide the three equation of motion using a velocity time graph v=U+at s =ut+ 1/2at² 2 V² 2= u²2+2as