Question

Derive the kinematic equations of motion for constant acceleration

Answer 1

Let a particle moves from point A with intial speed u , with constant acceleration particle reaches the point B where its speed became v ,time taken by particle from A to B is t .
See attachment,
slope of v- t time graph = Acceleration
(v - u)/t = a
v = u + at [ formula] -----(1)

Now, To find displacement, find area inclosed by v-t graph
so, displacement = area incosed by v- t graph
= 1/2 [ v + u] t
= 1/2 (u + at + u)t [ from equation (1)
= ut + 1/2 at²
Hence , S = ut + 1/2 at² [formula]-----(2)

Now, v = u + at
squaring both sides,
v² = u² + a²t² + 2uat
= u² + 2a[ut + 1/2at²]
= u² + 2aS [ from equation (2)
Hence, v² = u² + 2aS [ formula]

Answer 2

:

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]