Question

derive the third equation of motion : v=u+at

Answer 1

BC =v and AO =GC =u

BC=BG+GC ,hence we have

v=BG+u

-: BG=v-u

as per graph , the uniform acceleration of object ,

a=change in velocity/time taken =BG/AG = DA/OC = v-u/t

: v-u =at or v = u+at

BY ANALYTICAL METHOD :

by defination of acceleration , we know that

Acceleration ,a =change in velocity /time taken

= final velocity - initial velocity /time

= v-u /t

at =v-u or v=u+ at

Answer 2

HLO FRIEND YOUR ANSWER IS

the total distance as travelled by the object in time T is given by area under V -t graph .

therefore ,

distance travelled,S = Area OPQRT Obviously OPQRT is is a trapezium having area

$(\frac{op + qt}{2} ) \\ \times ot$

but OP=u ,

QT=v

and

OT=t

Therefore,

putting the value, distance travelled,

S=area of trapezium

=

$( \frac{op + qt}{2} ) \times ot = (\frac{u + v}{2}) \times t$

we know that at=v-u i.e.,

$t \ = \frac{v -u }{a}$

on substituting this value of t,

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

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

HOPE THIS IS HELPFUL TO YOU