Question

derivation of equations of motion by graphical method elementary idea of uniform circle motion​

Answer 1

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★Uniform Circular Motion

When a body moves in a circle, it is called circular motion.

When a body moves in a circular path with uniform speed its motion is called uniform circular motion.

Equation of motion by graphical method.

❒Derivation of v=u +at

Initial velocity u at A =OA

Velocity changes from A To B in time t(uniform acceleration a)

Final Velocity v=BC

$◑⇝BC=BD+DC \\ \\ ◑⇝v=BD+AO \\ \\ ◑⇝v=BD+u$

Slope of velocity time graph is equal to acceleration a.

$◑⇝a= BD\\ \\ ◑⇝a = \frac{BD}{AD} \\ \\ ◑⇝a = \frac{BD}{t} \\ \\◑⇝ BD = at \\ \\ ◑⇝v=u+at$

❒Derivation for $s = ut + \frac{1}{2} a {t}^{2}$

The distance travelled by the body is given by area of the space between velocity time graph AB and time axis OC , which is equal to area of figure OABC.

Distance travelled=Area of figure OABC

= Area of rectangle OADC + Area of triangle ABD

$◑⇒(OA \times OC) + \frac{1}{2} \times AD \times BD) \\ \\ ◑⇒(u \times t) + ( \frac{1}{2} \times t \times at) \\ \\ ◑⇒S = ut + \frac{1}{2} \times at$

❒Derivation of v = u+ 2as

The distance travelled by body in time t is given by area of figure OABC (which is a trapezium)

$s= Sum \: of \: parallel \: sides \times height \div 2 \\ \\◑↣ s = Area \: of \: trapezium \: OABC \\ \\ ◑↣s= (OA+OB) \times \frac{OC}{2} \\ \\ ◑↣at = v – u \\ \\ ◑↣s=(u+v) \times \frac{t}{2} \\ \\◑↣ v=u + at \\ \\◑↣ t=\frac{(v- u)}{2a} \\ \\ ◑↣s= (u+v) \times \frac{(v- u)}{2a} \\ \\ ◑↣2as = {v}^{2} - {u}^{2} \\ \\ ◑↣{u}^{2} = {v}^{2} + 2as$

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Answer 2

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