# Motion:

is one of the most basic and fundamental activities in the universe. Stars, planets, comets, satellites are all in motion. Planet earth rotates as well as revolves as p[art of its incessant motion within the solar system.

Motion is often described as 'relative.' This implies that in a given situation the same activity may appear as stationary to one observer while another observer might see it differently. For eg…..let's take the case of an aeroplane speeding along the runway….about to take off!! For two co-passengers sitting next to each other…..each one appears stationary with respect to the him/herself & to the plane. This is because both are moving together within the same plane at the same speed. However, both find themselves moving with respect to the airstrip/ground outside the plane with considerable speed. Likewise, an outside observer finds the people within the plane moving very fast.

**Concepts of distance & displacement: (9 April 2010)**

Distance is a scalar quantity while displacement is a vector quantity. A scalar quantity only has magnitude whereas a vector quantity has both magnitude and direction. If I say….a car travels a distance of 100km (magnitude)…I am referring to distance. If I say…a car travels 100km (magnitude) to the West (direction)…I am referring to displacement.

Concepts of distance and displacement when coupled with time give us the quantities of SPEED & VELOCITY.

**Uniform and Non-Uniform Motion: (13 April 2010)**

When a body covers/travels equal distance in equal intervals of time…we say it is performing uniform motion. The word 'uniform' implies similarity….sameness. A school 'uniform' ensures that all kids in the school have uniformity in their dress-up for a given day. In the same way, in physics uniform motion is typical of a body that displays sameness in its motion over a period of time.

Lets take up an example….I am sure you must have heard of Usain Bolt…the fastest man in the world. If not, then please wake up!!! He is the current 100m sprint World & Olympic Gold winner….and that too with extraordinary records!! When Bolt starts off the race from the blocks, his speed is zero at the very beginning. But over the next 100m he builds it up in such a way that at the finish line his speed is maximum. This is a typical example of non-uniform motion…..Usain covers unequal distances in equal intervals of time. For every 10m that he runs, he picks up his speed. Thus, for the stretch 20 to 30m the speed that he has is less than what he has for the 60 to 70m stretch.

A marathon runner who has to cover a long distance is much more likely to run in stretches where he maintains uniform motion. In these stretches his speed remains the same…and he covers equal distances in equal intervals of time.

From the above we can conclude….that uniform motion is non-accelerated motion, whereas non-uniform motion is accelerated motion. Why?? We shall discuss this a little later.

Nikhil

**15 April 2010**

Concept of **Average Speed** helps us to get an idea of how fast or slow we are traveling over a longish stretch of a road (say)……. I drove to my parents' house in Dehra Dun about a week ago. Starting very early….4am!!….I was able to cover good distance till about 7am. My parents house is about 245km from mine in Ghaziabad. In the first 3 hours….from 4am - 7am….I covered about 175km. I finally reached my destination at 9am. **Can you guys tell me what my average speed was??**

Average Speed = Total distance/Total time

As regards **Average Velocity**….which is a vector quantity…we generally use the concept of initial velocity (u) and final velocity (v).

Average Velocity = (u + v)/2

Velocity….being a vector…is dictated by both magnitude and direction. Suppose a car is moving at **45km/hr**(magnitude) **westwards** (direction). If either of the two……'45km/h' or 'westward' changes….we say that the velocity has changed. Thus, if we now have 45k/h moving southwards….we can say that velocity has changed….which implies that **acceleration** is coming into play!!

When velocity changes uniformly over a period of time…we call it **uniform acceleration**. Suppose….

- at t=0sec, u=5m/s…and for t=10sec, v=25m/s. Then rate of change of velocity = (v-u)/t = (25-5)10 = 20/10 = 2m/s
^{2} - at t=10sec, u=25m/s…and for t=20sec, v=45m/s. Then rate of change of velocity = (45-25)/10 = 20/10 = 2m/s
^{2}

In both cases we find that the rate of change of velocity remains uniform, or conserved. This implies **uniform acceleration** or that the rate of change of velocity is not changing.

**Now coming to acceleration:**

Suppose u=initial velocity; v=final velocity; t=time taken for u to become v

Then, change of velocity = (v-u) in t seconds

Applying unitary method….we have:

In t seconds…….change in velocity = (v-u)

Therefore, in 1 sec……..change = (v-u)/t

**This per second change in velocity is referred to as acceleration. Its units:**

a = (v-u)/t……Now putting the SI units below:

= (m/s)/s

= m/s^{2}

**Thus, the SI unit for acceleration becomes m/s ^{2}**

Acceleration a = (v-u)/t

If we multiply both sides by t we get:

at = v-u

**or v=u+at……….this is the First Equation of Motion. We have just derived it using basic definitions and the unitary method**

Nikhil

**16 April 2010**

We start with graphical representation of motion from next week. Please complete your notes by the weekend.

Nikhil

**19 April 2010**

Graphical representation of data helps us to visualise information better. The 'Manhattan' in Cricket very effectively represents runs scored per over in an ODI or a T20 match.

The same way we represent various physical quantities like distance, speed, velocity etc versus time on the X-Y co-ordinates. This representation is very effective in helping us understand how motion is being described in actuality.

To start with, let us take a DISTANCE Vs TIME Graph. Refer the fig below:

- We have distance (meters) on y axis and time (seconds) on the x axis
- Here, we describe the motion of a car that starts from rest
- At t
_{1}= 1 sec let d_{1}= 1m; t_{2}= 4 sec d_{2}= 4m - We have plotted these points on the x-y co-ordinates using the basics of co-ordinate geometry
- Now, consider triangle ABC. As per the definition of speed, the slope of the distance-time graph gives us speed. This slope refers to tan (theta), which means (Perpendicular)/(Base) within triangle ABC

Thus, we have:

Slope of the graph = P/B

= BC/AC

= (BX - CX)/(OX - OY)

Putting in the values, we get:

Slope of the graph = Speed from A to B = (4m - 1m)/(4s - 1s) = 3m/3s = 1m/s

This value of speed has been obtained by the graphical method. The same is also obtained if we simply apply the formula Speed = D/T to the data:

At t_{1}= 1 sec let d_{1}= 1m; t_{2}= 4 sec d_{2}= 4m

The above graph represents **uniform motion**…..as can be seen from the car covering equal distances in equal intervals of time. The graph belopw represents **non-uniform motion**…..covering unequal distances in equal intervals of time.

In this figure….the body covers 5km in the first 15mins but then covers 10km (goes from 5km to 15km) in the next 15mins (from the 15th to the 30th minute)….thereby covering unequal distances in equal intervals of time.

Nikhil

**21 April 2010: Velocity - Time Graphs:**

Refer Fig 1 and 2 below. Fig 1 represents a V-T graph depicting constant velocity….of 10m/s….with AB. This implies uniform velocity….meaning velocity remains constant even as time movers forward.

An interesting aspect about V-T graphs is this that we can find out the distance (or displacement) in a given period of time by finding the area of the geometric figure formed between the graph, the x axis, the y axis and the perpendicular from the graph to the x axis. In Fig 1, we can find displacement after 5 seconds as follows:

Area of rectangle with length L = 5seconds and breath B = 10m/s…..

= L x B = 5s x 10m/s = 50m. In the calculation, the 2 seconds (s) cancel out and we have meter (m) left…which represents displacement. Thus, Area within this V-T graph gives us displacement. If we were dealing with a Speed - Time graph, the same area would represent Distance. Can you tell why??

Now consider Fig 2……V-T graph with velocity changing with time. Any change in velocity represents acceleration. If this change is uniform (as in Fig 2 above; that is velocity changes equally for equal intervals of time) the acceleration produced can be called CONSTANT ACCELERATION.

This acceleration can be obtained as the Slope of Graph AB (Can you remember we did the same with the D-T graph earlier? Check out if you have forgotten). This Slope is also tan theta…where theta represents angle BAC.

Thus, Slope of AB = Acceleration = BC/AC

= (20m/s-10m/s)/(4s-2s)

= (10m/s)/2s

= 5m/s^{2}

Also, displacement in the t=4s is given by the area of the graph as explained with regard to Fig1.

Displacement = Area of triangle ABC = 1/2 x base x height

= 1/2 x AC X BC

= 1/2 x 2s x 10m/s

= 10m

**Thanks to Shilpi of Class IX A for the above figure. I welcome more students to make neat figures so that these can be shared through my notes.**

Nikhil

**26 April 2010**

Figures 8.7 (a) and (b) both represent non-uniform velocity. Can you tell why??

In Fig 8.7(a) we see that as times passes, velocity is decreasing. However, this decrease is not very uniform….as can be seen by the dots just above and below the graph line. Thus, even though the reduction in velocity is happening, this reduction is not the same for equal intervals of time.

In Fig 8.7(b) the change in velocity with time is random….velocity sometimes increases and sometime decreases. We thus have acceleration and retardation occurring one after the other…..another eg of non-uniform velocity.

**Derivation of Equations of Motion using graphs**

Graphs are great ways to visualize motion. Here, we use them to derive the 3 most fundamental equations of motion.

Let us consider a bus having initial velocity = u m/s at time = 0 sec. After 't' seconds it becomes final velocity = v

These initial and final states of motion are represented by points A and B on graph Fig 8.8….a V-T graph

Let the origin of this graph be 'O'

Join AB to get the graph representing uniform acceleration.

From B, drop a perpendicular to x axis, to meet it at C (this coincides with final time = t sec)

From A drop a perpendicular to BC to meet BC at D. Let BE be a perpendicular on the y axis

Now, let us consider triangle ABD:

Acceleration of this motion = slope of AB = BD/AD

= (BC-DC)/AD

= (v-u)/t

Thus, a = (v-u)/t

or at = v-u

**or v=u + at ……………….The First Equation of Motion………….Equation (1)**

Suppose, during this time 't' the bus has travelled a distance = 's' meters

Therefore, s = Area of trapezium OABC

= 1/2(SUM OF PARALLEL SIDES)X DISTANCE BETWEEN THEM

= 1/2(AO+BC)x OC

= 1/2(u+v)t

or s = 1/2(u+v)t

or s = 1/2(u+u+at)t……….putting v=u+at from Eqn 1 above

or s = 1/2t(2u+at)

**or s = ut + 1/2at ^{2}………………….The Second Equation of Motion……….Equation (2)**

Now, we derive the 3rd Equation as follows:

s = 1/2(u+v)t…..from derivation of 2nd Equation above, and

t = (v-u)/a……… from derivation of 1st Equation above, we get

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

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

or as = 1/2(v^{2}-u^{2})

or 2as = v^{2}-u^{2}

**or v ^{2} = u^{2} + 2as…………….3rd Equation of Motion (3)**

**1 May 2010**

**Circular Motion**

Refer the fig 8.9

Look how as we keep increasing the sides of a rectangle (a) it finally becomes a circle (d)!!

If an athlete runs around a square track he has to turn 4 times. In a hexagon (b) he turns 6 times, octagon (c) 8 times and finally, as we keep increasing the sides infinitely, the polygon eventually becomes a circle!!!! Thus, in a circle, the athlete is constantly changing his direction, every fraction of a second. So, even if his speed remains the same, the direction changes every moment….and so we can say that its speed is constant (uniform motion) but velocity changes (due to changing direction), thereby resulting in accelerated motion.

Circular motion is thus a very interesting case of UNIFORM MOTION (due to constant speed) + ACCELERATION (due to changing velocity). Uniform Circular Motion is therefore ACCELERATED MOTION.

We get: v = velocity = displacement/time

= circumference of circle/time

= (2 x pie x r)/t meters/second

There are many examples of this type of motion. Can you think of some??

Students, its time now to start our first Assignment:

**Assignment 1** - Click here to view and download

Nikhil