## Kerala Plus One Physics Notes Chapter 3 Motion in a Straight Line

**Mechanics**

It is the branch of Physics, which deals with the study of motion of physical bodies. Mechanics can be broadly classified into following branches.

- Statics: It is the branch of mechanics, which deals with the study of physical bodies at rest.
- Kinematics: It is the branch of mechanics, which deals with study of motion of physical bodies without taking into ac count the factors, which causes motion.
- Dynamics: It is the branch of mechanics, which deals with the study of motion of physical bodies taking into account the factors which causes motion.

**Rest and Motion Rest (നിശ്ചലാവസ്ഥ). **An object is said to be at rest if it does not change its position w.r.t its surroundings with time, eg:, a person sitting on a chair.

** Motion (ചലനം).**

An object is said to be in motion if it changes its position w.r.t its surroundings with time, eg., a bird flying in air, train moving.

Note: Rest and motion are relative, eg., a person sitting in a moving train is at rest w.r.t his fellow passengers but is in motion w.r.t to a person outside the train. Similarly a person sitting in his office is at rest w.r.t earth but is in motion w.r.t other planets.

**Types of motion**

- Translatory motion. Here particles of the object have same displacement, eg., motion of a car along a road.
- Rotatory motion. Here all the particles in the object moves along a circular path, eg., rotation of earth on its axis.
- Oscillatory motion. To and fro motion, eg,, motion of the pendulum of a wall clock. Apart from these, motion can be classified into one dimensional (eg., motion of freely falling body), two dimensional (eg., motion of planets around sun), three dimensional (eg., kite flying) motions.In this chapter we will study only one dimensional motion (straight line).

** Point object (സൂക്ഷ്മ ****വി)**

If the distance travelled by the body is very large compared with its size, the size of the body may be neglected. The body under such a condition may be taken as a point object. The point object can be represented by a point.

eg.,

- The length of bus may be neglected compared with the length of the road it is running.
- The size of planet is ignored compared with the size of the orbit in which it is moving.

**Position, Path length and Displacement**

**1. Position**

Position of an object is always defined with respect to some reference point which we generally refer to as origin. To define the change in position we have two physical quantities

**2. Path length or Distance**

It is the length of the actual path travelled by a body between its initial and final position.

**3. Displacement (സ്ഥാനാന്തരം)**

It is the change in the position of an object in a fixed direction. It is the shortest (or straight line) path measured in the direction from initial point to final point.

Consider an object moving from A to B along the path ACB. The distance travelled is the length of the path ABC and the displacement is AB.

Differences between distance and displacement

Distance | Displacement |

Scalar Quantity | Velocity Quantity |

For a moving particle distance can never be zero or negative. | displacement can be zero, positive or negative. |

For a moving particle, distance can never decrease with time. | Decrease in displacement with time means that the body is moving towards the initial position. |

The distance is either equal to or greater than displacement but never less than displacement | Magnitude of displacement less than or equal to distance |

**Uniform motion**

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

Position-time graph of an object in uniform motion.

**Speed and Velocity**

**Speed (വേഗത)**

The distance travelled by an object per second is called speed. It is also defined as the rate of change of position.

Speed has only magnitude, but no direction, so it is a scalar quantity.

The SI unit of speed is ms^{1}.

Dimensional formula is [MLT^{1}].

**Average speed (ശരാശരി വേഗത)**

It is the ratio of the total distance travelled by the object to the total time taken.

If a body covers distances s_{r} s_{2}, s_{3},…. with speeds v_{r} v_{2}, v_{3} then its average speed is

Special case. If s_{1} = s_{2} = s, which means the body covers equal distances with different speeds, then

**Velocity (പ്രവേഗം)**

It is the rate of change of displacement of an object.

Velocity has both magnitude and direction, so it is a vector quantity.

The SI unit and dimensional formula of velocity is same as that of speed

**Uniform velocity (സമപ്രവേഗം)**

If a body covers equal displacements in equal intervals of time, it is said to be moving with uniform velocity.

Note: For an object having uniform speed along a straight line, the magnitude of displacement = distance covered. Thus, for a straight line motion. uniform speed and uniform velocity are same.

**Variable velocity**

If a body’s speed or direction of motion or both changes with time, then it is said to be moving with variable velocity.

Consider an object with uniform velocity . Let time be measured from the instant, the object is at O. Let A and B be the positions of the object at the instants t_{1 }and t_{2 }respectively then

**Average velocity (ശരാശരി പ്രവേഗം)**

It is defined as the ratio of its total displacement to the time taken for the displacement.

Note: Average velocity is used for a body travel-ling with variable velocity.

The average velocity can be positive or negative depending upon the sign of the displacement. It is zero if the displacement is zero.

**Position-time graph for an object moving with**

**Instantaneous Speed and Velocity**

If velocity of a body increases with time, we say the body is accelerating [remember the accelerator pedal of cars]. The rate of change of velocity of an object is called its acceleration.

Note: The speedometer of a vehicle shows the instantaneous speed at any time

Question 1.

Can a body have a constant speed and still have a varying velocity?

Answer:

Yes. A particle in uniform circular motion has a constant speed but varying velocity because of the change in its direction of motion at every point

**Accelaration**

If velocity of a body increases with time, we say the body is accelerating [remember the accelerator pedal of cars]. The rate of change of velocity of an object is called its acceleration.

**Uniform Accelaration**

The acceleration of an object is said to be uniform acceleration, if its velocity changes by equal amounts in equal intervals of time.

**Average acceleration**

For an object moving with non-uniform acceleration, average acceleration is defined as the ratio of the total change in velocity of the object to the total time interval taken.

If v_{1} and v_{2} are the velocities of an object at instants t_{1} and t_{2} respectively, then its average acceleration is,

**Instantaneous acceleration (ക്ഷണ ത്വരണം).**

Acceleration of an object at a given instant of time is called instantaneous acceleration.

So, acceleration is the first order derivative of velocity and second order derivative t of position w.r.t time.

**Negative acceleration**

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

Question 2.

Is it possible for a body to be accelerated without speeding up or slowing down? if so, give an example 2.

Answer:

Yes. An object in uniform circular motion is accelerating but its speed neither decreases nor increases.

**Position- time graph**

Position-time graph for motion with

**Kinematic equations for uniformly accelerated motion**

Let an object move with uniform acceleration ‘a’ along a straight line OX with origin at O. Let the object reach the points A and B at time t_{1} and t_{2} with displacements x_{1} and x_{2 }respectively. Let v_{1} and v_{2} be the velocities of the object at A and B respectively. t=o t=t,

**1. Velocity-time relation (സമയ പ്രവേഗബന്ധം )**

**2. Position-time relation (സമയ-സ്ഥാന ബന്ധം )**

Let v

_{av}be the average velocity of the object during the motion from A to B.

If x_{t} = x_{0} is the displacement and v = u is the velocity at t= 0 and x = x be the diplacement and v_{2} =v be the velocity at

**3. Position – velocity relation (പവേഗ-സ്ഥാന ബന്ധം).**

**Position-time graphs (സമയ-സ്ഥാന ഗാഫ്) **Position-time graph for a stationary object (object at rest)

Let the object be stationary at position x

_{(t)}= x

_{0}from the origin

**Position-time graph for uniform motion along a straight line (uniform velocity)**

Thus the slope of the position-time graph gives the velocity of the object.

Question 3.

Position-time graph for uniformly accelerated motion

Thus the slope of the position-time graph for uniform accelerated motion gives the instantaneous velocity of the object.

**‘Velocity – time graphs (സമയ- പവേഗ ഗ്രാഫ്)**

**Velocity – time graph for uniform motion**

As the velocity remains constant with time in uniform motion, we see that the graph is a straight line parallel to the time axis.

Area under the v – t graph between t_{1 }and t_{2}

= Area of rectangle PQRS

= QR x SR = v(t_{2}– t_{1} )

= velocity x time

= displacement

So, the area under the velocity-time graph gives the displacement of the object in the given time interval.

**Velocity-time graph for uniformly accelerated motion. **

As the velocity changes by equal amount in equal intervals of time we see that the graph is a straight line inclined to the time axis

Slope of the velocity – time graph AB

So the slope of the velocity-time graph gives the acceleration of the object.

So the slope of the velocity-time graph gives the acceleration of the object.

**Area under the velocity -time graph**

Consider an object moving along a straight line AB with uniform acceleration ‘a’. Let v0 and v be the velocities of the body at times 0 and t respectively. The velocity-time graph of the motion looks like the below

= average velocity x time interval

= distance travelled in time t

So area under the velocity-time graph gives the distance travelled by the object in the given time interval.

Note: Make it clear that in the above, distance and displacement are same as the motion is along a straight line.

So the slope of the velocity-time graph gives the acceleration of the object.

Question 3.

A burglar’s car had a start with an acceleration of 2 ms^{[-2]}. A police vigilant party came after 5 seconds and continued to chase the burglar’s car with a uniform velocity of 20 ms’^{1}. Find the time in which the police van overtakes the burglar’s car.

Answer:

Suppose the police van overtakes the burglar’s car in time t second after it starts chasing. Distance covered by police van in t seconds = 20 t. Distance covered by burglar’s car in (t + 5) seconds,

s = u (t +5)+1/2 a (t+ 5)^{2 }= 0(t + 5) + 1/_{2} x 2(t + 5)^{2} = (t + 5)^{2 }Now 20t = (t + 5)^{2 }or t^{2}– 10t +25 = 0

or (t – 5)^{2} = 0

∴ t = 5s.

**Free-Fall**

An object released (near the surface of earth) is accelerated towards the earth. If air resistance is neglected, the object is said to be in free fall.

The acceleration due to gravity (g) near the surface of earth is 9.8 m/s^{2} (g).

Note: Free-fall is a case of motion with uniform acceleration.

Suppose a body is projected vertically upward from a point A with velocity u.

If we take upward direction as positive,

- At time t, its velocity, v = u – gt
- At time t, its displacement from A is given by, h = ut – 1/2 gt
^{2 } - Its velocity when it has a displacement ‘h’ is given by, v
^{2}= u^{2}– 2gh - When it reaches the maximum height from A, its velocity v = 0. This happens

when t = u/g. The body is instantaneously at rest at the highest point. - The maximum height reached,
- Total time to go up and return to the point of projection, T-2u/g

Question 4.

Two balls are thrown simultaneously, A vertically upwards with a speed Of 20 ms^{-1}^{ }from the ground, and B vertically downwards from a height of 40 m with the same

speed and along the same line of motion. At what points do the two balls collide?

Take g = 9.8 ms^{-2}.

Answer:

Suppose the two balls meet at a height of x metre from the ground after time t s from

the start.

For upward motion of ball A:

u = 20 ms^{1}, g = -9.8 ms^{-2 }s = ut +1/2 gt^{2 }x = 20t-1/2x 9.8t^{2 }= 20 t- 4.9 t^{2} ……… (i)

For downward motion of ball B:

40- x = 20 × t + ^{1}/2 ×9.8 t^{2 }= 20 t + 4.91^{2} ……. (ii)

Adding (i) and (ii)

40 = 40t or t= 1 s

From (i), x = 20×1 -4.9 × (1)^{2} =15.1 m

Hence the two balls will collide after 1 sat a height of 15.1 m from the ground.

**Relative Velocity:**

This is the velocity of one body with respect to another body [Actually the difference

of their velocities). As velocity is a vector,we should check the directions too. So two bodies with same velocities (same speed and direction) will have their relative

velocities zero.

Relative velocity of an object A with respect to another object B is the time rate at

which the object A changes its position with respect to the object B.

; where are the velocities of object A and B. v indicates the addition of negative of velocity of B to the velocity of A.

a. If v_{B}= v_{A –} v_{b}VA 0. Then,

x_{B}(t)— x_{A}(t)=x_{B}(0)— x_{A}(0).

Therefore. the two objects stay at a constant distance .

(x_{B}(0)— x_{A}(O)) a part,and their position-time graphs are straight lines parallel to each other. The relative velocity v_{BA }is zero in this case.

b. If v_{A} > v_{B}, v_{B} – v_{A} is negative.

One graph is steeper than the other and they meet at a common point.

c. If v_{B} > v_{A}, v_{B} – v_{A} is positive.This is the case when body B overtakes body A

d. Suppose v_{A} and v_{B} are of opposite directions. In this case, the magnitude of v_{BA }

or v is greater than the magnitude of velocity of A or that of B.

Note: The effect of relative velocity is seen when you sit in a train which is not moving, but you feel like you are in motion when a train passes by the side in opposite direction. Then there is relative velocity!

Question 5.

A car A moving at 10 ms^{-1} on a straight road, is ahead of car B moving in the same direction at 6 ms’^{1}. Find the velocity of A relative to B and vice versa.

Answer:

Here υ_{A} = 10 ms^{-1}, v_{g} = 6 ms^{-1 }Velocity of A relative to B,

υ_{AB} = υ_{A}-υ_{B}=10-6=4ms-^{1 }Positive velocity indicates that the driver of car B sees car A moving ahead from him at the rate of 4 ms^{-1}.

Velocity of B relative to A,

υ_{BA=}υ_{B – }υ_{A} = 6 -10= – 4ms^{ -1 }Negative velocity indicates that the driver of car A (when we looks back) sees the car B lagging behind at the rate of 4 ms^{-1}.

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