Question

State & proof all the equation of motion with graph &the 6numerical on three equation of motion but from not book

Answer 1

The three equation of motions are......

Let an object is moving with uniform acceleration.

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 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+△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.

Equation for Distance Velocity Relation: Third equation of Motion:

The distance covered by the object moving with uniform acceleration is given by the area of trapezium ABDO

Therefore,

Area of trapezium ABDOE

=12×(sum of parallel sides+distance between parallel sides)=12×(sum of parallel sides+distance between parallel sides)

⇒ Distance (s) =12(DO+BE)×OE=12(DO+BE)×OE

⇒s=12(u+v)×t⇒s=12(u+v)×t ----(iii)

Now from equation (ii) a=v−uta=v-ut

∴t=v−ut∴t=v-ut ----(iv)

After substituting the value of tt from equation (iv) in equation (iii)

⇒s=12(u+v)×(v−u)a⇒s=12(u+v)×(v-u)a

⇒s=12a(v+u)(v−u)⇒s=12a(v+u)(v-u)

⇒2as=(v+u)(v−u)⇒2as=(v+u)(v-u)

⇒2as=v2−u2⇒2as=v2-u2

⇒2as+u2=v2⇒2as+u2=v2

⇒v2=u2+2as⇒v2=u2+2as

The above expression gives the relation between position and velocity and is called the third equation of motion.

While solving the problems related to velocity, distance, time and acceleration following three points should be considered....

I Hope this answer will help u...