Question

with the help of a graph derive second law of motion
( with picture plz)

Answer 1

Motion
Equation for Velocity-Time relation by graphical method: First equation of Motion
Let an object is moving with uniform acceleration.

Velocity Time Graph
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 velocity
Time taken

⇒a=
v-u
t

⇒a=
OC-OD
t

=
DC
t

⇒at=DC -----(ii)

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

v=at+u
⇒v=u+at
Above equation is the relation among initial vlocity (u), final velocity (v), acceleration (a) and time (t). It is called first equation of motion.



Equation for distance –time relation
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+ Area of ADOE

⇒s=
1
2

×AB×AD+(OD×OE)
⇒s=
1
2

×DC×AD+(u+t)
[Since, AB=DC]

⇒s=
1
2

×at×t+ut
⇒s=
1
2

×at×t+ut
[∵ DC=at]

⇒s=
1
2

at2+ut
⇒s=ut+
1
2

at2
The above expression gives the distance covered by the object moving with uniform acceleration. This expression is known as second equation of motion.