E1.3 Kinematics (Equation of Motion in Straight Line):

For an object moving in a straight line, displacement, velocity, acceleration and time are taken are related by simple equation called Kinematical Equations (or equations of motion in straight line).

I. Analytical Treatment: 

a) Distance covered with Uniform Velocity:                                              $s$ = $ut$ 

b) Velocity of a Uniformly accelerated body after time (t):                       $v$ = $u$ + $at$ 

c) Distance covered by a uniformly accelerated body in time (t):     $s$ = $ut + 1/2at^2$ 

d) Velocity of Uniformly accelerated body after covering a distance (S):  $v^2$ - $u^2$ = $2as$ 

e) Distance traveled in $n^{th}$ second:                          $s_{n^{(th)}}$ = $u + 1/2 a(2n - 1)$

II. Graphical Treatment:

a) v = u + at 

Acceleration = slope of the velocity-time graph AB 

∴ a = $\frac {BM}{AM}$ = $\frac {BM}{ON}$ = $\frac {BN - MN}{ON}$ = $\frac {v - u}{t}$ 

or, $v - u = at$ 

⇒ $v = u + at$     .................... (i)

b) $s = ut + \frac{1}{2}at^2$

Acceleration = slope of the velocity-time graph AB 

∴ a = $\frac{BM}{AM}$ = $\frac{BM}{t}$; ⇒ $BM$ = $at$ 

Now, distance traveled by object in time t, 

s = area of trapezium OABN = area of rectangle OAMN + area of triangle ABM 

   = OA * ON + $\frac{1}{2}OB$ * AM = ut + $\frac{1}{2}at$ * t = ut + $\frac{1}{2}at^2$ 

Therefore, s = ut + $\frac{1}{2}at^2$            .............. (ii)

c) $v^2$ = $u^2$ + 2as 

From the velocity - time graph: 

s = area OABN 

= $\frac{1}{2}(OA + NB).AM$ 

= $\frac{1}{2}(OA + NB). \frac{AM}{BM}.BM$ 

= $\frac{1}{2}(u + v). \frac{1}{BM/AM}.(BN - MN)$ 

= $\frac{1}{2}(u + v)\frac{1}{a}(v - u)$ 

= $\frac{v^2 - u^2}{2a}$ 

∴ $v^2 - u^2 = 2as$        .......................... (iii)

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