Question

Cannu explain the EQUATION OF MOTION BY GRAPHICAL METHOD (explain with graph)

Answer 1

Answer:

this is the answer for your question

thank you

Answer 2

To Derive v = u + at By Graphical Method

Let the initial velocity (u) at point A.

Its velocity changes at uniform rate from A to B in time (t).

Its final velocity (v) which is equal to BC

Let OC = t

Now, Initial Velocity of the body (u) = OA

Finish Velocity of the body (v) = BC

Then, BC = BD + DC

Therefore, v = BD + DC

So, v = BD + OA (Since DC = OA)

Therefore, v = BD + u -------(1st eq.)

We know that,

Slope of velocity-time graph is equal to acceleration (a)

Then, a = Slope of line (AB)

a = BD/AD

a = BD/t

Therefore, BD = at

In eq. 1st

v = BD + u

v = at + u (First eq. solved)

To derive s = ut + 1/2at^2 By Graphical Method

Suppose, the body travels the distance (s) in time (t)

We know that,

Area of the space between the velocity-time graph AB and the time axis OC which is equal to the area of the figure OADC

Thus, Distance travelled = Area of figure OADC or Area of rectangle OADC + Area of triangle ABD-----(1st eq.)

Area of rectangle OADC = OA × AC

= u × t

= ut-----(2nd eq.)

Area of triangle ABD = 1/2 × Area of rectangle AEBD

= 1/2 × AD × BD

= 1/2 × t × at

= 1/2at^2-----(3rd eq.)

From eq. 1st,

s = Area of rectangle OADC + Area of triangle ABD

s = ut + 1/2at^2 (2nd eq. solved)

To derive v^2 = u^2 + 2as By Graphical Method

Suppose, distance travels (s) by a body in time (t)

We know that,

Distance = Area of space OABC (which is a trapezium)

Area of trapezium OABC, s = Sum of parallel × h/2

s = (OA + BC) × OC / 2

s = (u + v) × t / 2

2s = (u + v) × t

From the equation of motion 1,

v = u + at

at = v - u

t = v - u / a

Therefore,

2s = (u + v) v - u / a

2as = uv - u^2 + v^2 - uv

v^2 = u^2 + 2as (3rd eq. solved)