**CBSE NOTES CLASS 9 SCIENCE CHAPTER 8**

**MOTION**

**Motion in a Straight Line**

Motion of objects along a straight line is called **rectilinear motion**. The object can move forward or backward, up or down etc.

**Distance**

The distance travelled by an object, without considering the direction, is called the **path length**. It is a scalar quantity.

**Displacement**

Displacement is the shortest distance from the initial to the final position of an object.

It is a vector quantity.

A displacement vector represents the length and direction of the straight line joining the initial and final point.

The magnitude of displacement is less than or equal to the actual distance (path length) travelled by the object in the given time interval.

Displacement ≤ Distance

The two quantities are equal only if the object does not change its direction during the course of its motion.

- Find the distance and displacement if an object moves O-C-B-A-B.

**Uniform Motion**

If an object moving along a straight line covers equal distances in equal intervals of time, it is said to be in **uniform motion **along a straight line.

**Average speed **

**Average speed **is defined as the total distance travelled divided by the total time interval during which the motion has taken place

$$\mathrm{Average\; speed\; =}\frac{\mathrm{Total\; distance\; travelled}}{\mathrm{Total\; time\; taken}}$$

Or v = $\frac{\mathrm{s}}{\mathrm{t}}$

Speed is a scalar quantity.

Its SI unit is meter/sec. It is never negative.

**Velocity**

The rate of change of position or displacement of an object, in a particular direction with respect to time is called velocity.

It is equal to the displacement covered by an object a unit time.

In case the velocity of the object is changing at a uniform rate, then average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time.

**Average velocity** is defined as the change in position or displacement divided by the time intervals, in which the displacement occurs,

$${\mathrm{v}}_{\mathrm{a}\mathrm{v}}\mathrm{}=\mathrm{}\frac{\mathrm{Inital\; velocity\; +\; Final\; velocity}}{\mathrm{Total\; time\; taken}}\mathrm{}=\mathrm{}\frac{\mathrm{u\; +\; v}}{\mathrm{\Delta \; t}}$$

Also average velocity

$${\mathrm{v}}_{\mathrm{a}\mathrm{v}}\mathrm{}=\frac{\mathrm{displacement}}{\mathrm{time\; interval}}=\frac{{\mathrm{s}}_{2}-{\mathrm{s}}_{1}}{{\mathrm{t}}_{2}-{\mathrm{t}}_{1}}$$

- The SI unit for velocity is ms
^{–1}. It is a vector quantity. - It may be negative, positive or zero.
- When a body moves in a straight line and in the
**same direction**, the average speed is equal to the magnitude of average velocity.

**Acceleration **

The rate of change of velocity of an object with respect to time is called its acceleration.

$$\mathrm{acceleration}=\frac{\mathrm{change\; in\; velocity}}{\mathrm{time\; taken}}$$

If the velocity of an object changes from an initial value u* *to the final value v* *in time from t_{1} to t_{2} , the acceleration *a *is,

$$\mathrm{a}\mathrm{}=\mathrm{}\frac{\mathrm{v}-\mathrm{u}}{{\mathrm{t}}_{2}-\mathrm{}{\mathrm{t}}_{1}}$$

It is a vector quantity. Its SI unit is meter/sec^{2}.

It may be positive, negative or zero.

**Positive acceleration **

If the velocity of an object increases with time, its acceleration is positive, that is x and a have same direction.

**Negative acceleration**

If the velocity of an object decreases with time, its acceleration is negative. The negative acceleration is also called retardation or deceleration.

**D-T Diagram for Uniform Motion**

**D-T Diagram for Non-Uniform Motion**

**V-T graphs**

- Area under the v-t curve represents the displacement over a given time interval.
- The area under the a-t curve represents the change in velocity (v-u).

**Equations of Uniformly Accelerated Motion along a Straight Line**

1. v = u + *at* …(Velocity-Time Relation)

2. s = ut + $\frac{1}{2}$*a t*^{2} …(Position-Time Relation)

3. v^{2} – u^{2} = 2*as* …(Position-velocity Relation)

**Proofs (Graphical method)**

1. From the diagram,

OA = CD = u,

OE = CB = v,

OC = AD = t

CB = CD + DB

DB = CB – CD = v – u

Now since,

$$\mathrm{a}\mathrm{}=\frac{\mathrm{change\; in\; velocity}}{\mathrm{Time\; taken}}=\frac{\mathrm{D}\mathrm{B}}{\mathrm{O}\mathrm{C}}$$

$$\mathrm{a}\mathrm{}=\mathrm{}\frac{\mathrm{v}\mathrm{}\u2013\mathrm{}\mathrm{u}}{\mathrm{t}},\mathrm{}\mathrm{}$$

Therefore, $\mathrm{v}=\mathrm{u}+\mathrm{a}\mathrm{t}$

2. Displacement s = Area under the v-t curve between instants 0 and t

= Area of rectangle OADC + Area of triangle ADB

= OC × AO + $\frac{1}{2}$ × AD × DB

= ut + $\frac{1}{2}$ × t × (v-u)

Now since v – u = at (from first equation), we have

s = ut + $\frac{1}{2}$ × t × at,

or **s **= ut + $\frac{1}{2}$ at^{2}

3. Displacement s = Area under the v-t curve between instants 0 and t

= Area of trapezium OABC

$$=\frac{\mathrm{O}\mathrm{A}+\mathrm{B}\mathrm{C}}{2}\times \mathrm{}\mathrm{O}\mathrm{C}=\frac{\mathrm{v}+\mathrm{u}}{2}\times \mathrm{t}$$

Now since t = $\frac{\mathrm{v}-\mathrm{u}}{\mathrm{a}}$ (from first equation), we have

$$\mathrm{s}\mathrm{}=\frac{\mathrm{v}+\mathrm{u}}{2}\times \mathrm{t}=\frac{\left(\mathrm{v}+\mathrm{u}\right)\left(\mathrm{v}-\mathrm{u}\right)}{2\mathrm{a}}$$

$$\mathrm{s}\mathrm{}=\frac{{\mathrm{v}}^{2}-\mathrm{}{\mathrm{u}}^{2}}{2\mathrm{a}}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$$

$$\Rightarrow \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathrm{v}}^{2}-\mathrm{}{\mathrm{u}}^{2}=2\mathrm{a}\mathrm{s}$$

**Uniform Circular Motion**

When an object moves in a circular path with **uniform speed**, its motion is called **uniform circular motion.**

$$\mathrm{v}\mathrm{}=\frac{2\mathrm{\pi}\mathrm{r}}{\mathrm{T}}$$

Where, **T** is the time taken by the object in circular motion to complete one round.

Although the speed is constant, the velocity is changing in direction.

The direction of velocity is along a straight line tangential to the circular path.

There is a centripetal force and centripetal acceleration acting on the object towards the centre of the circular path.

$${\mathrm{a}}_{\mathrm{c}}=\frac{{\mathrm{v}}^{2}}{\mathrm{r}}\mathrm{}$$

And

$${\mathrm{F}}_{\mathrm{c}}=\mathrm{}\frac{{\mathrm{m}\mathrm{v}}^{2}}{\mathrm{r}}$$