Question

obtain the equation 2as=v2-u2 from the equation v=u+at and s=ut+1/2at2​

Answer 1

Answer:

$\large\bold\red{{v}^{2}-{u}^{2}=2as}$

Explanation:

Given,

$v = u + at$

Squaring both the sides,

We get,

$= > {v}^{2} = {(u + at)}^{2} \\ \\ = > {v}^{2} = {u}^{2} + {a}^{2} {t}^{2} + 2uat \: \: \: \: .......(1)$

Also,

Given that,

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

Multiplying both the sides with 2a,

We get,

$= > 2as = 2aut + {a}^{2} {t}^{2} \: \: \: ..........(2)$

Now,

Subtracting eqn (2) from (1) ,

We get,

$= > {v}^{2} - 2as = {u}^{2} \\ \\ = > \bold{{v}^{2} - {u}^{2} = 2as}$

Hence, Proved.

Answer 2

equations:

v=u+at                                                                                   ...........1

$s=ut+\frac{1}{2} at^{2$                                              ...........2

in eq.1

v=u+at

v-u=at

t=(v-u)/a

sustituting this value in eq. 2

$s=u(\frac{v-u}{a} )+\frac{1}{2} a(\frac{v-u}{a} )^{2} \\s=\frac{vu-u^{2} }{a} +\frac{\frac{1}{2} (v^{2} +u^{2} -2vu)}{a} \\as=\frac{2vu-2u^{2} +v^{2} +u^{2} -2vu}{2} \\2as=v^{2} -u^{2}$

hence proved