Derive the third equation of motion by graphical method.

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Solution

Step 1: Finding 3rd equation of motion using the graph method
Let us consider the initial velocity of a body be $u$ , and final velocity of a body be $v$ , the acceleration of the body be $a$ and the time is taken by body be $t$ .
We know that the first equation of motion is $v = u + a t$
Also, we know that the second equation of motion is $s = u t + \frac{1}{2} a t^{2}$ [Here s is distance covered]
Step 2: Finding displacement
Displacement covered during the time interval $t_{2} - t_{1} = A r e a A B t_{2} t_{1}$
$S =$ Area of triangle $A B A^{1}$ + Area of rectangle $A A^{1} t_{2} t_{1} = A r e a o f t r a p e z i u m$
$S = \frac{1}{2} \times s u m o f p a r a l l e l s i d e s \times P e r p e n d i c u l a r d i s \tan c e$
Sum of parallel sides = $O D + O E = u + v$
Perpendicular distance = $t_{2} - t_{1} = t$
Then $S = \frac{1}{2} \times (v + u) t . . . . . (i)$
Step 3: Finding 3rd equation of motion
From the first equation of motion $t = \frac{v - u}{a}$
Substituting equation (i) we get
$S = \frac{1}{2} (v + u) \frac{(v - u)}{a}$
After solving we get the third equation of motion is $v^{2} - u^{2} = 2 a S$ .
Hence, $v^{2} - u^{2} = 2 a S$ is the third equation of motion.

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Similar questions

Derive the second equation of motion by a method other than the graphical method.

Derive the following equation of motion by the graphical method :

$v^{2} = u^{2} + 2 a$

where the symbols have their usual meanings.

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