Question

Derive
the
equations of motion by
Graphical method?

Answer Fast.

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Answer 1

Answer:

Let the initial velocity of the object = u

Let the object is moving with uniform acceleration, a.

Let object reaches at point B after time, t and its final velocity becomes, v

Draw a line parallel to x-axis DA from point, D from where object starts moving.

Draw another line BA from point B parallel to y-axis which meets at E at y-axis.

Let OE = time, t

Now, from the graph,

BE = AB + AE

⇒ v = DC + OD (Since, AB = DC and AE = OD)

⇒ v = DC + u (Since, OD = u)

⇒ v = DC + u ------------------- (i)

Now, Acceleration (a) =Change in velocityTime taken=Change in velocityTime taken

⇒a=v−ut⇒a=v-ut

⇒a=OC−ODt=DCt⇒a=OC-ODt=DCt

⇒at=DC⇒at=DC -----(ii)

By substituting the value of DC from (ii) in (i) we get

v=at+uv=at+u

⇒v=u+at⇒v=u+at

Above equation is the relation among initial vlocity (uu), final velocity (vv), acceleration (a) and time (t). It is called first equation of motion.

Equation for Distance -Time Related

Distance covered by the object in the given time ‘t’ is given by the area of the trapezium ABDOE

Let in the given time, t the distance covered by the moving object = s

The area of trapezium, ABDOE

= Distance (s) = Area of △ABD+△ABD+ Area of ADOE

⇒s=12×AB×AD+(OD×OE)⇒s=12×AB×AD+(OD×OE)

⇒s=12×DC×AD+(u+t)⇒s=12×DC×AD+(u+t)

[Since, AB=DCAB=DC]

⇒s=12×at×t+ut⇒s=12×at×t+ut

⇒s=12×at×t+ut⇒s=12×at×t+ut

[∵ DC=atDC=at]

⇒s=12at2+ut⇒s=12at2+ut

⇒s=ut+12at2⇒s=ut+12at2

The above expression gives the distance covered by the object moving with uniform acceleration. This expression is known as second equation of motion.

Explanation:

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