ACCELERATION  & EQUATIONS OF MOTION

ACCELERATION  & EQUATIONS OF MOTION
 GRAPHICAL TREATMENT: KINEMATICAL EQUATIONS OF MOTION OF A BODY WITH UNIFORM ACCELERATION
Consider a particle moving with initial velocity 'u' and uniform acceleration 'a'. Suppose 'v' is its velocity after 't' seconds. Let 'S' be its displacement in the time interval t. The velocity time graph is a straight line with positive slope. The graph is given by the line AB.

1) To show that v=u+at
The slope of velocity time graph gives the acceleration of the particle.
Here, slope = \(\tan \theta = \frac{{BC}}{{AC}} = \frac{{v - u}}{t} = a\)
     (or) v=u+at
2) To Show that S= \(ut+\frac{1}{2}a{t^2}\)
The area under the velocity time graph with X-axis (time axis) gives the displacement in the bounded time interval. Here the area bounded by the line AB with x-axis gives the displacement
S= Area of rectangle OACD + Area of triangle ABC
S=(OA) (OD)+\(1\over 2\)(AC)(CB)
=(u)(t)+(t)(v-u)  \(\left[ {\therefore a = \frac{{u - v}}{t}} \right]\)
= ut +(t)(at)   
=\(\therefore S = ut + \frac{1}{2}at\)  
Note : General Method to show that  \(S = ut + \frac{1}{2}a{t^2}\)
When the particle is moving with uniform acceleration,
verage velocity =\(\frac{{u + v}}{2}\)
 Displacement = (Average Velocity) (time)
\(\therefore S = \left( {\frac{{u + v}}{2}} \right)t\)
\(\therefore S = \left( {\frac{{u + u + at}}{2}} \right)t\)      (V=u+at)
\(\therefore S = ut + \frac{1}{2}at\)
Note : To show that \({v^2} - {u^2} = 2aS\)
V=u+at        ..V-u = at...............(1)
\(S = \left( {\frac{{u + v}}{2}} \right)t\)    \(\therefore v + u = \frac{{2S}}{t}\)   ..............(2)
 Using (1), (2)
(v-u) (v + u)=(at)\(\frac{{2S}}{t}\)=2aS
\(\therefore {v^2} - {u^2} = 2aS\)