derive the equation for position time relation for an object travelling under uniform acceleration on a straight path.
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Answers
S = ut +
1
2
at2
Where,
S = displacement
u = initial velocity
a = acceleration (must be constant),
t = time.
This equation DOES NOT relate to final velocity.
Analytical Proof
We know that,
v = u + at
Also,
v =
dS
dt
‘S’ is the displacement.
Therefore, equating both,
dS
dt
= u + at
S∫0 dS = t∫0 (u + at) . dt
S = ut∫0 dt + a t∫0 t . dt
S = ut +
1
2
at2
Graphical Proof
Following is a v - t graph displaying constant acceleration. (Slope of the curve is constant)
Curve Constant
At t = 0 seconds, the particle’s velocity is u m/s.
At t = t seconds, the particle’s velocity is v m/s.
Area under the curve of v − t graph gives displacement.
Now,
S = Area of Rectangle + Area of Triangle
S = (u - 0) × (t - 0) +
1
2
(v - u)(t - 0)
S = ut +
1
2
(v - u)(t)
(Substituting v − u = at, from first equation of motion and rearranging the terms)
S = ut +
1
2
at2