Question

Derive 2nd equation of motion with graph

Answer 1

sry but i can't tell the graph
theory :
let us consider that the object had travelled a distance s in time t under uniform acceleration a. the distance travelled is the area enclosed within oabc in the velocity time graph ab. thus distance travelled is given by ...
s = area of oabc ( which is a trapezium )
= area of triangle abd + area of rectangle oadc
= 1 ÷ 2 × ad × bd + oc × ao
= 1÷2 × oc × bd + oc × ao

a is v - u ÷ t
bd ÷ t
therefore bd = at

bd is st
oc is t
ao is u
bc is v
let us put these in the equation
s = ut + 1 ÷ 2 at square
hence proved

Answer 2

 Second equation of motion:

s = ut + 1/2 at^2

sol.

Let the distance travelled by the body be “s”.

We know that

Distance = Average velocity X Time

Also, Average velocity = (u+v)/2

.: Distance (t) = (u+v)/2 X t …….eq.(1)

Again we know that:

v = u + at

substituting this value of “v” in eq.(2), we get

s = (u+u+at)/2 x t

=>s = (2u+at)/2 X t

=>s = (2ut+at^2)/2

=>s = 2ut/2 + at^2/2

or s = ut +1/2 at^2