Motion

The relations among velocity, acceleration, time and distance of an object in motion are called EQUATIONS OF MOTION.

There are mainly three types of Equations of motion. These equations are

`v=u+at` ------------(i)

`s= ut + 1/2\ at^2` --------- (ii)

`2as = v^2-u^2` -------------- (iii)

Where `u` = initial velocity

`v`= final velocity

`s` = distance

`a` = acceleration

And `t` = time.

Equation (i) describes the velocity-time relation.

Equation (ii) describe the position time relation.

And equation (iii) represents the relation between position and velocity.

These three equations of motion can be derived by graphical method.

Equations of Motion by Graphical Method

Let consider the velocity time graph of an object that is moving with uniform acceleration.

In this graph,

Initial velocity =`u (u!=0)` [At point A]

Final velocity `=v` [At point B]

The initial velocity `u` increases to `v` in time `t`

That is time `=t`

Now, two perpendiculars BC and BE are drawn from point B on the time and velocity axes respectively.

Thus, Initial velocity is represented by OA, and the final velocity is represented by BC.

And time interval `t` is represented by OC.

Equation for Velocity Time relation by Graphical Method

BD = BC – CD

This represents the change in velocity in time interval `t`.

Now, let draw AD parallel to OC.

Thus, from the graph

BC = BD + DC

`=>BC = BD + OA`

Now, after substituting BC `=v` and OA `=u`, we get

`v = BD + u`

`=>BD = v-u` -----------(iv)

Now, we know that acceleration (`a`) of the object is the change in velocity in unit time.

Thus, Acceleration (`a`) `=(text{Change in velocity})/(text{time taken})`

`=>a = (BD)/(AD)`

`=>a = (BD)/(OC)`

After substituting OC = `t`, we get

`=>a = (BD)/t`

`=>BD = at` ------------(v)

Now, after substituting the value of BD in equation (iv) we get

`v= u+at` ------------- (vi)

Where `u`= initial velocity, `v`=final velocity, `a` = acceleration and `t`= time.

This is known as velocity equation of motion for velocity time relation.

By knowing any three value among `u, v, a` and `t` the fourth can be calculated using equation for velocity time relation.

Equation for Position Time Relation by Graphical Method

Let the object has travelled a distance `s` in time `t` with uniform acceleration `a`.

Thus, in the given graph, distance travelled by the object is obtained by the area enclosed within OABC, which is a trapezium.

∴ `s` = area of OABC

= area of the rectangle OADC + area of the triangle ABD

`=OAxxOC+1/2 (ADxxBD)`

Now, since OA = `u` and OC = AD =`t` and BD = `at`

Thus, after substituting the values of OA, OC, AD and BD, we get

`s = uxxt+1/2(txxat)`

`=>s=ut+1/2at^2` --------- (vii)

This equation is called equation for Position Time relation.

Thus, by knowing three among `s,u,t` and `a`, fourth can be calculated.

Equation for Position Velocity Relation

The distance travelled by the object moving with uniform acceleration `a` in time `t` is given by the area enclosed within the trapezium OABC given in the graph.

That is, distance `s` = area of the trapezium OABC

= `1/2`(sum of parallel sides)× distance between parallel sides.

`= 1/2(OA+BC)xxOC`

Now, after substituting OA `=u`, BC `=v` and OC `=t`, we get

`s=1/2(u+v)t` ----------- (viii)

Now, we know from velocity time relation that

`v=u+at`

`=>at = v-u`

`=t = (v-u)/a`

Thus, after substituting the value of `t` from above equation in equation (viii), we get

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

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

[∵` (u+v)(u-v)=v^2-u^2`]

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

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

This equation is known as equation for position relation velocity.

Problem Example (1): A bus gets a velocity of 54 km/h in 5 minutes starting from rest. Find its acceleration assuming that acceleration is uniform and find the distance covered to attain this velocity.

Solution:

Given, Initial velocity `(u)` = 0 [Because bus starts from rest]

Final velocity `(v)` = 50 km/h

`=(54xx1000)/(60xx60)\ m//s`

`=54000/3600 = 1.5\ m//s`

And, given, time = 5 minute

`= 5xx60 = 300\ s`

Or, `t = 300\ s`

Then, Acceleration `a` = ?

And distance `s` =?

Calculation of acceleration

Now, we know that, `v = u+at`

`=> 1.5\ m// s = 0 + a xx 300\ s`

`=> 1.5\ m// s = a xx 300\ s`

`=> a = (1.5\ m// s)/(300\ s)`

`=> a = 0.005\ m\ s^(-2)`

Calculation of distance covered while achieving this velocity

We know that, `s = ut + 1/2 at^2`

`=> s = 0 xx 300\ s + 1/2 xx 0.005\ m \ s^(-2) xx (300\ s)^2`

`=> s = 1/2 xx 0.005\ m\ s^(-2) xx 90000\ s^2`

`=> s = 0.005\ m\ s^(-2) xx 45000\ s^2`

`=> s = 225\ m`

Thus, acceleration (`a`) = 0.005 m s^–2, and distance (`s`) = 225 m Answer.

Problem Example (2): A train gets a velocity of 36 km/h in 2 minutes. Find the acceleration of the train and distance covered to achieve the given velocity.

Solution:

Given,

As initial velocity is not given, thus assuming that train starts from rest.

Thus, initial velocity, `u` = 0

Final velocity, `v = 36\ km// h`

Since, the SI unit of acceleration is m/s/s, thus this final velocity is to be converted into m/s

Thus, 36 km/h `=(36 xx1000\ m)/(60xx60\ s)`

`=(36000\ m)/(3600\ s) = 1\ m\ s`

Thus, Final velocity, `v = 1\ m//s`

Given, time `t` = 2 minute = 2 × 60 = 120 s

Calculation of Acceleration (a)

We know that, `v = u+at`

`=>1\ m//s = 0 + a xx 120\ s`

`=>axx120\ s = 1\ m//s`

`=>a = (1\ m//s)/(120\ s)`

`=>a = 0.0083\ m\ s^(-2)`

Calculation of distance covered to achieve the final velocity

We know that, `s = ut + 1/2 at^2`

`=>s = 0 xx 120\ s + 1/2 xx 1/120\ m\ s^(-2) xx (120\ s)^2`

`=> s = 1/2 xx 1/120\ m\ s^(-2) xx 120\ s xx 120 s`

`=>s = 1/2 xx 120 m = 60\ m`

Thus, acceleration `= 1/120 \ m\ s^(-2)` or `0.0083\ m\ s^(-2)` and distance = 60 m Answer

Example Problem (3): A train is running at a velocity of 36 km/h and accelerates uniformly to 72 km/h in 10 second. Find the acceleration and distance covered to achieve the new velocity.

Solution:

Given, Initial velocity, `u` = 36 km/h

`=( 36xx1000\ m)/(60xx60\ s)`

`=(36000\ m)/(3600\ s)`

`=>u = 10\ m//s`

And given, final velocity, `v = 72\ km//h`

`=(72xx1000\ m)/(60xx60\ s)`

`=(72000\ m)/(3600\ s)`

`=>v= 20\ m//s`

And, given, time `t` = 10 second.

Thus, acceleration (a) and distance (s)=?

Calculation of acceleration

We know that, `v = u +at`

`=> 20\ m//s = 10\ m//s + a xx 10\ s`

`=>axx 10\ s = 20\ m//s-10\ m//s`

`=>axx10\ s = 10\ m//s`

`=>a = (10\ m//s)/(10\ s)`

`=>a = 1\ m\ s^(-2)`

Calculation of distance

We know that `s=ut+1/2 at^2`

`= 10\ m//s xx 10\ s+1/2 xx 1\ m\ s^(-2)xx(10\ s)^2`

`=100 m + 1/2 xx1\ m\ s^(-2)xx100\ s^2`

`=100\ m + 50\ m`

`=>s = 150\ m`

Thus, acceleration (a) = 1 m s^–2 and distance (s) = 150 m Answer

Example Problem (4): A bus is running at a speed of 72 km/h. After applying brakes by driver bus stops after 10 second. Find the acceleration and distance covered by bus before getting stopped.

Solution:

Given, Initial velocity, `u` = 72\ km/h

`=(72xx1000\ m)/(60xx60\ s)`

`=(72000\ m)/(3600\ s)`

`=>u=20\ m//s`

Because bus stops after braking, thus Final velocity `u=0`

Time, `t` = 10 second

Calculation of acceleration

We know that, `v=u+at`

`=>0 = 20\ m//s+axx10\ s`

`=>axx10\ s= -20\ m//s`

`=>a = (-20\ m//s)/(10\ s)`

`=>a=-2\ m\ s^(-2)`

Calculation of distance

We know that, `s = ut+1/2at^2`

`=>s = 20\ m//sxx10\ s+1/2 xx (-2\ m\ s^(-2))xx(10\ s)^2`

`=200\ m + 1/2 xx (-2\ m\ s^(-2))xx100\ s^2`

`=200\ m-100\ m`

`=>s = 100\ m`

Thus, acceleration (a) = –2 m s^–2 and distance (s) = 100 m Answer

Uniform Circular Motion

The motion of an object along a circular path is called circular motion. Motion of an athlete along a circular path, motion of hands of clock, motion of earth, motion of blades of a fan, motion of top, etc. are the examples of uniform circular motion.

Acceleration changed when direction of object is motion is changed. While moving along a circular path the direction of object is changing continuously, hence motion of an object along a circular path is called accelerated motion also.

Velocity of an object moving in circular motion

Let the radius of the circle along which object is moving `=r`

Therefore, circumference of the circle `=2pir`

Now, if object completes one round in time `t`,

We know that, velocity `(v) = (text{distance})/(text{time})`

Here, in the case of circular motion, distance = circumference of circle `=2pir`

And time `=t`

Thus, velocity `(v) = (2pir)/t`

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

Where, `v` = velocity, `r`= radius of the circular path, and `t` = time taken to complete one rotation.