**This article is formulated according to the 3rd chapter in the NCERT Syllabus of 11th Std Physics OR 1st PUC Physics.**

## Mechanics

Mechanics is the oldest and fundamental branch of physics and it is the study of the state of rest as well as the state of motion of an object under the action of force. The study of mechanics is broadly classified into (i) Statics and (ii) Dynamics

## Statics

It deals with bodies at rest under the action of system of force.

## Dynamics

It deals with motion of a body under the action of force. Dynamics is again divided into (a) Kinematics and (b) Kinetics.

## Kinematics

It deals with the description of motion without reference to the cause of motion.

## Kinetics

It deals with what moves and what causes motion. Some of the terms used in describing motion are given below.

## Particle

A particle is ideally just a piece or quantity of matter, having no linear dimensions but only position and mass.

## Event

An event is a physical process that occurs at a point in space and at an instant of time.

## Observer

A person or equipment which can locate, record, measure and interpret an event is called an observer.

## Frame of reference

It is the reference in which an observer sits and makes the observations. In order to specify the position, we need to use a reference point and set of axes. The choice of set of axes in a frame of reference depends on the situation.

## Motion

Motion is change in position of an object with time.

## Rectilinear motion

Motion of objects along a straight line. Ex: A car moving along a straight road, A freely falling body.

## Rest

A body is said to be at rest when it does not change its position with time.

## Path length

It is the actual distance covered by a body in time π‘. It is also called as distance travelled.

- Path length is a scalar quantity.
- SI unit of path length is “πππ‘ππ” (π). Dimensions are π
^{0}πΏπ^{0}. - Path length depends on the actual path.
- Path length is always positive.

## Displacement

It is the shortest distance between the initial point and final point.

- It is vector quantity.
- SI unit of displacement is “πππ‘ππ”(π). Dimensions are π
^{0}πΏπ^{0}. - Displacement may be positive, negative and zero.
- Magnitude of the displacement can never be greater than path length.
- When a body moves in straight line displacement is equal to path length.
- It is independent of the actual path travelled and it denoted by βπ₯.

## Difference between path length and displacement

Path length | Displacement |

It is the actual distance covered by a body in time t. | It is also called as distance travelled It is the shortest distance between the initial point and final point. |

Path length is a scalar quantity. | It is vector quantity. |

Path length is always positive. | Displacement may be positive, negative and zero. |

Path length is always greater than or equal to displacement. | Displacement is always less than or equal to path length. |

## Speed

Speed is defined as the rate of change of position of a particle.

πππππ = πππ‘β πππππ‘β/π‘πππ π‘ππππ = π₯/π‘

- Speed is a scalar quantity.
- Its SI unit is πππ‘ππ πππ π πππππ (ππ
^{β1}). Dimensions are π^{0}πΏπ^{β1}. - It is always positive.
- Speed gives no indication about the direction of motion of the particle.

## Average speed

The average speed of a particle in motion is defined as the ratio of the total path length to the total time taken.

π΄π£πππππ π ππππ = π‘ππ‘ππ πππ‘β πππππ‘β/π‘ππ‘ππ π‘πππ π‘ππππ

Instantaneous speed (speed)

It is defined as the limit of average speed as the time interval is infinitesimally small.

## Velocity

Velocity is defined as the rate of change of displacement of a body.

πππππππ‘π¦ = πππ πππππππππ‘/π‘πππ π‘ππππ

π = π/π

- Velocity is a vector quantity.
- SI unit is πππ π πππππ (ππ
^{β}). Dimensions are π^{0}πΏπ^{β1}. - Velocity may be positive, negative or zero.

## Average velocity

The average velocity of a particle in motion is defined as the ratio of total displacement to the total time taken.

π΄π£πππππ π£ππππππ‘π¦ = π‘ππ‘ππ πππ πππππππππ‘/π‘ππ‘ππ π‘πππ π‘ππππ

π£Μ
= (π₯_{2} β π₯_{1})/(π‘_{2} β π‘_{1})

π£Μ = βπ₯/βπ‘

## Instantaneous velocity

Velocity is defined as the limit of average velocity as the time interval βπ‘ becomes infinitesimally small. π = π₯π’π¦ _{βπβπ} (βπ/βπ) = π
π/π
t

- Instantaneous velocity is also called velocity.
- In a position-time graph, the instantaneous velocity at a point is the slope to the tangent drawn to the curve at that point.
- Instantaneous speed or speed is the magnitude of velocity.

## Uniform velocity

If equal changes of displacement take place in equal intervals of time is called uniform velocity.

Note: When a body moves with uniform velocity, neither the magnitude nor the direction of the velocity changes.

## Difference between speed and velocity

Speed | Velocity |

It is defined as the ratio of the path length to the time taken. | It is defined as the ratio of displacement to the time taken. |

Speed is a scalar quantity. | Velocity is a vector quantity. |

It is always positive. | Velocity may be positive, negative or zero. |

## Acceleration

It is defined as the rate change of velocity of a particle.

π΄ππππππππ‘πππ = πβππππ ππ π£ππππππ‘π¦/π‘πππ π‘ππππ

π = π β π_{π}π

- Acceleration is a vector quantity.
- SI unit is πππ‘ππ πππ π ππ’πππ ππ π πππππ (ππ β2) and dimensions are π0πΏπβ2.
- Since velocity is a quantity having both magnitude and direction, Acceleration may result from a change in magnitude or a change in direction or changes in both.
- Acceleration can be positive, negative, or zero.
- The negative acceleration is called retardation or deceleration.

## Average acceleration

It is defined as the total change in velocity divided by the total time taken.

πΜ
= (π£_{2} β π£_{1})/(π‘_{2} β π‘_{1}) = βπ£βπ‘

## Instantaneous acceleration

It is defined as the limit of the average acceleration as the time interval βπ‘ becomes infinitesimally small.

π = π₯π’π¦ _{β}_{π}_{β}_{π} βπ/βπ = π
π/π
π

## Uniform acceleration

If the velocity of a body changes by an equal amount in equal intervals of time, however small these time intervals may be, is called uniform acceleration.

**Graphical representation of motion**

## Graph

A diagrammatical representation of the variation of one quantity with respect to another quantity is called a graph.

## Position-time graph

It is a graph obtained by plotting the instantaneous positions of a particle versus time. The slope of the position time graph gives the velocity of the particle.

## Velocity-time graph

A graph of velocity versus time is called a velocity-time graph.

- The area under the v-t graph with the time axis gives the value of displacement covered in a given time.
- The slope of the tangent drawn on a graph gives instantaneous acceleration.

## Uses of velocity-time (v-t) graph / Significance of velocity-time (v-t) graph

- It is used to study the nature of motion.
- It is used to find the velocity of the particle at any instant in time.
- It is used to derive the equations of motion.
- It is used to find displacement and acceleration.

## Velocity time-graphs

## The kinematic equation for uniformly accelerated motion

For uniformly accelerated motion, we can derive some simple equations that relate displacement (π₯), time taken (π‘), initial velocity (π£0), final velocity (π£), and acceleration (π). These equations are called Kinematic equations for uniformly accelerated motion. The Equations are, (i) π£ = π£_{0} + ππ‘ (ii) π₯ = π£_{0}π‘ + 1/2 ππ‘^{2} (iii) π£^{2} = π£_{0}^{2} + 2ππ₯

## Derivation of equation of motion by graphical method

(i) π = π_{π} + ππ

Consider a particle in motion with initial velocity π£_{0} and constant acceleration π.

Let π£ be the final velocity of the body at time π‘.

From graph, slope = π΅πΆ/π΄πΆ = (π΅π· β πΆπ·)/π΄πΆ

But, πΆπ· = ππ΄ and π΄πΆ = ππ·

Slope = (π΅π· β πA)/ππ· = π£ β π£_{0}π‘

But, slope of v-t graph gives the acceleration.

π = (π£ β π£_{0})/π‘

ππ‘ = π£ β π£_{0}

π£ β π£_{0} = ππ‘

π = π_{π} + ππ

(ii) π = π_{π}π + π/πππ^{π}

Consider a particle in motion with initial velocity π£_{0} and constant acceleration π.

Let π£ be the final velocity of the body at time π‘. From graph, π·ππ πππππππππ‘ = π΄πππ π’ππππ π£ β π‘ ππππβ

π₯ = π΄πππ ππ π‘πππππ§ππ’π ππ΄π΅π·

π₯ = ππππ ππ βπ΄π΅πΆ + ππππ ππ ππππ‘πππππ ππ΄πΆπ·

π₯ = [ (1/2) Γ π΄πΆ Γ π΅πΆ] + [ππ· Γ ππ΄]

π₯ = (1/2)π‘(π£ β π£_{0}) + π‘π£_{0}

But, π£ β π£_{0} = ππ‘

π₯ = (1/2)π‘(ππ‘) + π£_{0}π‘

π₯ = (1/2)ππ‘^{2} + π£_{0}π‘

π = π_{π}π + (π/π)ππ^{π}

(iii) π^{π} = π_{π}^{π} + πππ

Consider a particle in motion with initial velocity π£_{0} and constant acceleration π. Let π£ be the final velocity of the body at time π‘. From the graph, π·ππ πππππππππ‘ = π΄πππ π’ππππ π£ β π‘ ππππβ

π₯ = π΄πππ ππ π‘πππππ§ππ’π ππ΄π΅π·

π₯ = (1/2) (ππ΄ + π΅π·)π΄πΆ

π₯ = (1/2) (π£_{0} + π£)π‘

But, π£ β π£_{0} = ππ‘ and π‘ = (π£ β π£_{0})/π

π₯ = (1/2)(π£_{0} + π£)(π£ β π£_{0}/π)

π₯ = (1/2)(π£^{2} β π£_{0}^{2})/π

2ππ₯ = π£^{2} β π£_{0}^{2}

π£^{2} β π£_{0}^{2} = 2ππ₯

π^{π} = π_{π}^{π} + πππ

Note: The set above equations were obtained by assuming that at π‘ = 0, the position of the Particle π₯ is 0 (zero). When at π‘ = 0, If the position of the particle is at π₯_{0}(πππ π§πππ), then the equations are, (i) π£ = π£_{0} + ππ‘ (ii) π₯ β π₯_{0} = π£_{0}π‘ + (1/2)ππ‘^{2} (iii) π£^{2} = π£_{0}^{2} + 2π(π₯ β π₯_{0}).

## Free fall

An object released near the surface of the earth is accelerated downward under the influence of the force of gravity. If the air resistance is neglected, then the motion of the body is known as free fall.

## Acceleration due to gravity

The acceleration produced in an object due to gravity is called acceleration due to gravity, denoted by π. Free fall is an example of motion along a straight line under constant acceleration.

- Acceleration due to gravity is always a downward vector directed toward the center of the earth.
- The magnitude of π is approximately 9.8ππ
^{2}near the surface of the earth. - Acceleration due to gravity is the same for all freely falling bodies irrespective of their size, shape, and mass.
- The distance traversed by a body falling freely from rest during equal intervals of time are in the ratio 1: 3: 5: 7: β¦ β¦ β¦. this is known as Galileoβs law of ODD numbers.

## Equations of motion under gravity

The motion of a freely falling body is in the Y-direction. If we take vertically upward as a positive Y-axis, acceleration is along the negative Y-axis, therefore π = βπ. Then, (i) π£ = π£_{0} β ππ‘ (ii) π¦ = π£_{0}π‘ β (1/2)ππ‘^{2} (iii) π£^{2} = π£_{0}^{2} β 2ππ¦

For a freely falling body the initial velocity, π£_{0} = 0. Then, (i) π£ = βππ‘ (ii) π¦ = β(1/2)ππ‘^{2} (iii) π£^{2} = β2ππ¦.

The π β π‘ graph, π£ β π‘ graph, and π¦ β π‘ graph to a body released from rest at π¦ = 0 are as shown.

**Note: (i) Stopping distance: **When breaks are applied to a moving vehicle, the distance traveled before stopping is called stopping distance.

π
π = βπ_{π}^{π}/ππ

It is an important factor for road safety, and it depends on initial velocity and deceleration (βπ).

(ii) **Reaction time**: When a situation demands immediate action, it takes some time before we really respond this time is called reaction time.

## Relative velocity

The relative velocity of body π΄ with respect to body π΅ is defined as the time rate of change of displacement of π΄ with respect to π΅.

**Explanation:** Consider two bodies A and B moving with constant velocity π£_{π΄} and π£_{π΅} respectively, along the positive X-axis. Let π₯_{π΄}(π‘) and π₯_{π΅}(π‘) be the position of π΄ and π΅ at any given instant of time π‘, then π₯_{π΄}(π‘) = π₯_{π΄}(0) + π£_{π΄}π‘, π₯_{π΅}(π‘) = π₯_{π΅}(0) + π£_{π΅}π‘.

The separation between π΄ and π΅ at time π‘ is,

π₯_{π΅}(π‘) β π₯_{π΄}(π‘) = π₯_{π΅}(0) β π₯_{π΄}(0) + (π£_{π΅} β π£_{π΄})π‘.

Here, π₯_{π΅}(0) β π₯_{π΄}(0) is the separation between π΄ and π΅ at π‘ = 0, and (π£_{π΅} β π£_{π΄}) is the time rate of change of relative velocity of π΅ with respect to π΄, denoted by π£_{π΅π΄}. Hence, π_{π©π¨} = π_{π©} β π_{π¨}.

Similarly, the velocity of A with respect to B is π_{π¨π©} = π_{π¨} β π_{π©} and it can be shown that π£_{π΄π΅} = βπ£_{π΅π΄}.

Case (1):

When two bodies move with the same velocity in the same direction, then π£π΄ = π£π΅ and π£π΄ β π£π΅ = 0 and π£π΄π΅ = π£π΅π΄ = 0, then two bodies appear at rest with respect to each other. In this case, the relative velocity is minimum.

Case (2):

When two bodies move in the same direction with different velocities, If π£_{π΄}>π£_{π΅ }then π£_{π΅π΄} = πππππ‘ππ£π and π£_{π΄π΅} = πππ ππ‘ππ£π.

Case (3):

When two bodies move in different velocities or the same velocities in opposite directions. The magnitude of the relative velocity of either of them with respect to the other is equal to the sum of the magnitude of their velocities. π£_{π΅π΄} = π£_{π΄π΅} = π£_{π΄} + π£_{π΅}.

In this case relative velocity is maximum.

## Problems

1) A car is moving along a straight line. It moves O to P in 18 second covering 360m and returns from P to Q in 6second by covering 120m. Calculate average velocity and average speed of a car in going (a) from O to P (b) from O to P and back to Q.

Given

ππ = 360π

ππ = 120π π

π = ππ β ππ = 360 β 120 = 240π

(a) Avg. velocity = πππ πππππππππ‘/π‘πππ

π£ = ππ/π‘ = 360/18 = 20ππ ^{β1}

Avg. speed = πππ π‘ππππ/π‘πππ

Avg. speed = 360/18 = 20ππ ^{β1}

(b) Avg. velocity = πππ πππππππππ‘/π‘πππ

π£ = ππ/π‘ = 240/24 = 10ππ ^{β1}

Avg. speed = πππ π‘ππππ/π‘πππ = (360 + 120)/24

Avg. speed = 480/24 = 20ππ ^{β1}

2) A car moving along a straight line takes 5 second to increase its velocity from 15ππ ^{β1} to 30ππ ^{β1}. What is the acceleration of the car? Also calculate the distance travelled by the car in 5 second.

Given π‘ = 5π , π£_{0} = 15ππ ^{β1}, π£ = 30ππ ^{β1}, π = ?, π₯ = ?

π£ = π£_{0} + ππ‘

30 = 15 + (π Γ 5)

5π = 30 β 15

π = 15/5 = 3ππ ^{β2}

π₯ = π£_{0}π‘ + (1/2)ππ‘^{2} π₯

= (15 Γ 5) + ((1/2) Γ 3 Γ 5^{2})

π₯ = 75 + (75/2)

π₯ = 75 + 37.5 = 112.5π

3) A car moving along a straight road increases its velocity from 10ππ ^{β1} to 30ππ ^{β1} in 4 seconds. Calculate (a) the acceleration of the car and (b) the distance travelled by the car in 4 seconds.

Given π£_{0} = 10ππ ^{β1}, π£ = 30ππ ^{β1}, π‘ = 4π , π = ?, π₯ = ?

π£ = π£_{0} + ππ‘

30 = 10 + (π Γ 4)

4π = 30 β 10

π = 20/4 = 5ππ ^{β2}

π₯ = π£_{0}π‘ + (1/2)ππ‘^{2}

π₯ = (10 Γ 4) + ((1/2) Γ 5 Γ 4^{2})

π₯ = 40 + (80/2)

π₯ = 40 + 40 = 80π

4) A truck moving along a straight highway with a speed of 20ππ ^{β1} is brought to rest in 10π . What is the retardation of the truck? How far will the truck travel before it comes to rest?

Given π£_{0} = 20ππ ^{β1}, π£ = 0ππ ^{β1}, π‘ = 10π , π =?, π₯ = ?

π£ = π£_{0} + ππ‘

0 = 20 + (π Γ 10)

10π = β20

π = β20/10 = β2ππ ^{β2}

π₯ = π£_{0}π‘ + (1/2)ππ‘^{2} = (20 Γ 10) + ((1/2) Γ (β2) Γ 10^{2})

π₯ = 200 β 100

π₯ = 100π

5) A player throws a ball upwards with an initial speed of 29.4ππ ^{β1} (a) What is the direction of acceleration during the upward motion of the ball? (b) What is the velocity and acceleration at the highest point of its path? (c) To what height does the ball rise and after how long does the ball returns to the playerβs hand?

Given π£_{0} = 29.4ππ ^{β1}

(a) The ball is moving under the gravity, so the direction of acceleration is vertically downwards and towards the centre of the earth.

(b) At the highest point, π£ = 0ππ ^{β1} and acceleration is equal to acceleration due to gravity, π = 9.8ππ ^{β2}

(c) π£^{2} = π£_{0}^{2} + 2ππ₯

0 = (29.4)^{2} + (2 Γ (β9.8) Γ π₯)

π₯ = (29.4)^{2}/(2 Γ 9.8) = 44.1π

π£ = π£_{0} + ππ‘

0 = 29.4 + ((β9.8) Γ π‘)

π‘ = 29.4/9.8 = 3π

Total time = time of ascent + time of decent = 3s +3s = 6s

6) A car is moving along a straight highway with a speed of 126kmph is brought to stop with in 200m. What is the retardation of the car and how long does it take for the car to stop?

Given π£_{0} = 126πππβ

π£_{0} = (126 Γ 1000)/3600ππ ^{β1} = 35ππ ^{β1}

π₯ = 200π

π£ = 0ππ ^{β1}

π = ? , π‘ = ?

π£^{2} = π£_{0}^{2} + 2ππ₯

0 = 352 + (2 Γ π Γ 200)

400π = β35 Γ 35

π = β (35 Γ 35)/400 = β3.06ππ ^{β2}

π£ = π£_{0} + ππ‘

0 = 35 + (β3.06) Γ π‘

π‘ = 35/3.06 = 11.42π

7) Two trains A and B of length 300m each are moving on two parallel tracks with a uniform speed of 54kmph in the same direction, with the train A ahead of B. The driver of train B decides to overtake A and accelerates by 2ππ ^{β2}. If after 25s, the guard of train B just brushes past the driver of A. What original distance between them?

Given π_{π΄} = 300π, π_{π΅} = 300π, π£_{π΄} = 54πππβ = 15ππ ^{β1}, π£_{π΅} = 15ππ ^{β1}, π_{π΄} = 0ππ ^{β2}, π_{π΅} = 2ππ ^{β2}, π‘ = 25π , Original distance, π = ?

Distance travelled by A in 25s, π₯_{π΄} = π£_{π΄}π‘ + ((1/2)π_{π΄}π‘^{2})

π₯_{π΄} = (15 Γ 25) + 0

π₯_{π΄} = 375π.

Distance travelled by B in 25s, π₯_{π΅} = π£_{π΅}π‘ + ((1/2)π_{π΅}π‘^{2})

π₯_{π΅} = (15 Γ 25) + ((1/2) Γ 2 Γ 25 Γ 25)

π₯_{π΅} = 375 + 625 = 1000

Distance travelled by B in 25s, π₯_{π΅} = π_{π΅} + π + π₯_{π΄} + π_{π΄}

1000 = 300 + π + 375 + 300

1000 = 975 + π

π = 1000 β 975 = 25π

8) The displacement (in metre) of a particle moving along x-axis is given by π₯ = 2π‘^{2}+3. Calculate (i) Average velocity between π‘ = 3π and π‘ = 5π . (ii) Instantaneous velocity at π‘ = 5 π and (iii) Instantaneous acceleration.

Given π₯ = 2π‘^{2}+3

(i)Avg. velocity = π‘ππ‘ππ πππ πππππππππ‘/π‘ππ‘ππ π‘πππ π‘ππππ When π‘_{1} = 3π ,

π₯_{1} = 2(3)^{2}+3 = 21π

When π‘_{2} = 5π , π₯_{2} = 2(5)^{2}+3 = 53π

Average velocity = (π₯_{2} β π₯_{1})/(π‘_{2} β π‘_{1})

π£Μ
= (53 β 21)/(5 β 3) = 32/2 = 16 ππ ^{β1}

(ii) Instantaneous velocity, π£ = ππ₯/ππ‘

π₯ = 2π‘^{2}+3

π£ = ππ₯/ππ‘ = 2(2)π‘ + 0 = 4π‘

When π‘ = 5π , π£ = ππ₯/ππ‘ = 4(5) = 20 ππ ^{β1}

(iii)Instantaneous acceleration, π = ππ£/ππ‘

π£ = 4π‘

π = ππ£/ππ‘ = 4ππ ^{β2}

9) Obtain equations of motion for constant acceleration using method of calculus.

(i) By definition, π = ππ£/ππ‘

ππ£ = πππ‘

Integrating both sides,

π£ β π£_{0} = π[π‘ β 0]

π£ β π£_{0} = ππ‘

π = π_{π} + ππ

(ii) Now, we have π£ = ππ₯/ππ‘ ππ₯ = π£ππ‘ Integrating both sides,

(π β π_{π}) = π_{π}π + (π/π)ππ^{π}

(iii) We have, π = ππ£/ππ‘ = (ππ£/ππ₯)(ππ₯/πt)

π = (ππ£/ππ₯)π£

πππ₯ = π£ππ£

Integrating on both sides,

π(π₯ β π₯_{0}) = (1/2)(π£^{2} β π£_{0}^{2})

2π(π₯ β π₯_{0}) = (π£^{2} β π£_{0}^{2})

π^{π} = π_{π}^{π} + ππ(π β π_{π})

## Important questions for exams.

One mark.

1) When will the magnitude of displacement equal to the path length?

2) Define average speed.

3) Define instantaneous speed.

4) Define instantaneous velocity.

5) Define average velocity.

6) Define acceleration.

7) What is retardation?

8) What is the acceleration of a body moving with uniform velocity?

9) What does the slope of position-time graph represent?

10) What does the slope of velocity-time graph represent?

11) Draw v-t graph for motion in uniform acceleration.

Two marks.

1) Distinguish between distance travelled and displacement of a particle.

2) Distinguish between speed and velocity.

3) Define uniform velocity and uniform acceleration.

4) What is position time graph? Draw π₯βπ‘ graph for an object at rest.

5) Draw position time graph for (a) a particle at rest,(b) a body moving with uniform velocity.

6) Draw the position time graph of a particle moving with a) Positive acceleration. b) Negative acceleration.

7) Draw v-t graph for body moving in uniform acceleration.

8) Define relative velocity. When will the relative velocity of two bodies be zero?

9) Define relative velocity with an example.

Three marks.

1) Write the significance of v-t graph.

2) Derive the equation π£ = π£_{0} + ππ‘ with usual notation by using v-t graph.

3) Define relative velocity. When does the relative velocity become maximum and minimum if two particles are moving along a straight line?

Five marks.

1) What is v-t graph? Derive the equation π£^{2} = π£_{0}^{2} + 2ππ₯ with usual notation by using v-t graph.

2) What is v-t graph? Derive the equation π₯ = π£_{0}π‘ + (1/2)ππ‘^{2} with usual notation by using v-t graph.

Additional Problems

1) The displacement of the particle moving along x-axis is given by π₯ = 3π‘^{3} β 5π‘^{2} + 1 where x is in metre and t is in second. Calculate (i) Instantaneous velocity at π‘ = 2 π (ii) Instantaneous acceleration at π‘ = 3 π

2) A body is thrown up with velocity of 78.4ππ ^{β1}. Find how high it will rise and how much time it will take to return to its point of projection.

3) A ball thrown vertically upwards and it reaches a height of 90π. Find the velocity with which it was thrown, and the height reached by the ball 7 second after it was thrown?

4) A body travelling with an initial velocity 36ππ ^{β1} comes to rest after travelling 90π. Assuming the retardation to be uniform, find its value. What time does it take to cover that distance?

5) A car is moving along a straight highway with a speed of 108ππβπ^{β1} is brought to stop with a distance 200π. What is the retardation of the car? And how long does it take for the car to stop?

6) A car travels a distance from A to B at a speed of 40ππβπ^{β1} and returns to A at the speed of 30ππ βπ^{β1}. What is the average speed for the whole journey?