Question

how to derive third equation of motion​

Answer 1

Answer:

here is your answer......

Answer 2

Answer:

third equation of motion is:

${v}^{2} - {u }^{2} + 2as$

area of trapezium OABC

$\frac {(oa + bc) \times oc}{2}$

$s \: = \frac{(u + v) \times t}{2}$

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

$s = \frac{ {v}^{2} \: - {u}^{2} } { {2a} }$

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

it is a velocity position relation