Class 9 Science

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) `=(text{Change in velocity})/(text{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.

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 `triangle ABD +` Area of ADOE

`=>s = 1/2 xx AB xx AD + (ODxxOE)`

`=>s = 1/2 xx DC xx AD + (u+t)`

[Since, `AB = DC`]

`=>s=1/2xx at xx t + ut`

`=>s=1/2xxatxxt+ut`

[∵ `DC=at`]

`=>s=1/2at^2+ut`

`=>s=ut+1/2at^2`

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

`=1/2xx(text{sum of parallel sides+distance between parallel sides})`

⇒ Distance (s) `=1/2(DO+BE) xx OE`

`=>s= 1/2(u+v) xxt` ----(iii)

Now from equation (ii) `a=(v-u)/t`

`:. t = (v-u)/t` ----(iv)

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

`=>s=1/2(u+v)xx((v-u))/a`

`=>s= 1/(2a)(v+u)(v-u)`

`=>2as=(v+u)(v-u)`

`=>2as = v^2-u^2`

`=>2as+u^2=v^2`

`=>v^2=u^2+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:

Motion of an object along a circular path is called circular motion. Since, on a circular path the direction of the object is changing continuously to keep it on the path, the motion of the object is called accelerated motion.

If the radius of circle is `r`

Therefore, circumference `=2pir`

Let time `t` is taken to complete one rotation over a circular path by any object

∴ Velocity `(v)=text{Distance}/text{time}`

`=>v=text{Circumference}/t`

`=>v=(2pir)/t`

Where, v = velocity, r = radius of circular path and t = time

Motion of earth around the sun, motion of moon around the earth, motion of a top, motion of blades of an electric fan, etc. are the examples of circular motion.

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