Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9

Class 9: Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9

The document Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9 is a part of the Class 9 Course Science - Short Notes Class IX.
All you need of Class 9 at this link: Class 9

Motion and Rest

  • An object that changes its position with respect to time is known to be in motion.
  • An object is said to be at rest when it does not change its position with respect to time. 
  • A reference point is required to describe a motion.
  • Example:
    (i) A car is moving at 60 km/hr. It means that the car is moving with respect to its surrounding. The reference point here is the surrounding.Car is in motion after the light is green whereas it is in rest as the light turns red
    Car is in motion after the light is green whereas it is in rest as the light turns red
    (ii) A person travelling by bus. The bus and the person are in motion with respect to the surroundings i.e. the plants, trees road, etc. The reference points here are the surroundings.
    But the person and the bus are at rest with each other because they are moving at the same speed and the man always remains at one point inside the bus. The reference point, in this case, is the bus for the man and vice-a-versa.

➢ Motion in a Straight Line

  • The point from where an object starts moving is treated as its reference point.
  • Distance: The total path length covered by the object from its reference point. Distance only has magnitude (magnitude = numerical value).
  • Displacement: The shortest distance covered by an object from its reference point. Displacement has both magnitude and direction.
  • The shortest distance is always a straight line dram from the reference point to the point of the end of the motion.
  • Example:
    Consider an object that moves from point O to C and then comes back to point A.Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9Distance covered by the object = OC + CA
    = 60 + 40 = 100 km
    Displacement, the shortest distance from the reference point, will be the distance from point O to the endpoint of end of motion i.e. point A, which is OA.
    Hence, displacement = OA
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9

➢ Uniform Motion

  • An object is said to be in uniform motion when it covers equal distances in equal intervals of times.
    Example: A car covered 40 km in each hour.
  • Graph for Uniform motion
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9Graph 1: Uniform Motion
    Graph 1: Uniform Motion

➢ Non–Uniform Motion

  • An object is said to be in non-uniform motion when it covers the unequal distance in equal intervals of time.
    Example: A man walking in a park.
  • Graph of Non-uniform Motion
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9Graph 2: Non-Uniform Motion 
    Graph 2: Non-Uniform Motion 

Speed and Velocity

  • Speed: Distance covered per unit time is called speed.
    Average speed = Total distance covered / Total time taken i.e. v = s/t
    where v = speed, s = distance travelled, t = time taken.
  • Velocity: Speed with direction is called velocity. Displacement per unit time is called velocity. The direction of velocity is the same as the direction of displacement or motion.
    Velocity = Displacement / Total time taken
    Average velocity =Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9 i.e. Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9
    where µ = initial velocity, v = final velocity.
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9Graph 3 - Distance vs Time at a constant velocity 
    Graph 3 - Distance vs Time at a constant velocity 

Try yourself:Which of the following statement is correct regarding velocity and speed of a moving body?
View Solution

Acceleration

  • The rate of change of velocity is known as acceleration.
    Acceleration = Change in velocity / Time taken i.e.Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9
    where v = final velocity, µ = initial velocity, t = time taken.
  • Graph
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9Graph 4 - Velocity vs Time at uniform acceleration
    Graph 4 - Velocity vs Time at uniform acceleration
  • The area under two points in a velocity-time graph gives the distance covered in that time.
  • The distance between time interval D (3hr) and C (5hr) is given by the area of the trapezium ABCD, i.e.
    Area of trapezium = (1/2) (b1 + b2) * h = (1/2) (AD + BC) * h
    = (1/2) * (30 + 50) * 2   (height = 5 – 3 (time period))
    = (1/2) * (80) * 2 = 80 km
  • But if the object moves with constant velocity and there is no acceleration, then velocity will be the same at all times.
    Velocity vs time graph at zero acceleration 
    Velocity vs time graph at zero acceleration 

Try yourself:A car of mass 1000 kg is moving with a velocity of 10 m/s. If the velocity-time graph for this car is a horizontal line parallel to the time axis, then the velocity of the car at the end of 25 s will be:
View Solution

Velocity-Time Equation

  1. v = u + at
  2. s = ut + (1/2) at2 
  3. 2as = v2 – u2

➢ Derivations of the Three Equations

(a) v = u + at

  • Consider an object moving at a uniform acceleration and forms the following graph as AB.Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9In the above graph:
    CD = Initial velocity = u, BD = Final Velocity = v, and OD = time = t
    ∵ BD = BC + CD
    ⇒ BC = BD – CD = v – u
    ∵ Acceleration = Change in velocity / Time taken
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9
    ∴ at = v – u
    i.e. u + at = v or v = u + at  
    Hence, proved or the first equation derived.
  • Deriving the above equation without the help graph
    Let, Change in velocity = Final velocity(v) – Initial velocity (u)
    Time taken = t and Acceleration = a
    ∵ Acceleration = Change in velocity / Time taken
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9
    i.e. at = v – u or v = u + at
    Hence, proved or first equation derived.

(b) s = ut + 1 / 2at2

  • Consider an object moving at a uniform acceleration and forms the following graph as AB.
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9In the above graph:
    The distance travelled in a velocity vs time graph is the area under two points.
     Distance travelled in the above graph between point A and B = Area of trapezium AODB
     s = Area of trapezium AODB = Area of rectangle AOCB + Area of triangle ACB
     s = OD * CD + (1/2) * AC * CB...(i)
     OD = time taken or ‘t’, CD = Initial velocity or ‘u’, AC = OD = time taken or ‘t’
     CB = DB – CD = (v – u)
    We know that:
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9 i.e. v – u = at
    ∴ CB = at
    Putting the values of OD, CD, AC, and CB in equation (i)
    ⇒ s = t * u + (1/2) * t * at
    ⇒ s = ut + (1/2) at2 
    Hence, proved or derived second equation.
  • Deriving the above equation without the help graph
    Average velocity = Total Distance covered / Time taken
    i.e.Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9 
    ⇒ s = vavg * t...(i)
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9
    Putting the value of Vavg in equation (i)
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9...(ii)
    We also know that: v = u + at
    Putting the value of v in equation (ii)
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9 
    ⇒ s = ut + 1 / 2at2 
    Hence, proved or derived the second equation.

(c) 2as = v2 – u2

  • Consider an object moving at a uniform acceleration and forms the following graph as AB.
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9In the above graph:
    The distance travelled in a velocity vs time graph is the area under two points.
    Distance (s) travelled in the above graph between point A and B = Area of trapezium AODB
    ⇒ s = Area of trapezium AODB = (1/2) * (b1 + b2) * h
    ⇒ s = (1/2) * (b1 + b2) * h...(i)
    From the graph: b1 = AO or ‘u’, b= DB or ‘v’ and ‘h’ = OD or ‘t’  
    Putting these values in equation (i)
    ⇒ s = (1/2) (AO + DB) * OD
    ⇒ 2s = (u + v) * t ...(ii)
    We also know that:
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9
    Putting the value of ‘t’ in equation (ii)
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9
    ⇒ 2as = v2 – u2
    Hence, proved or derived the third equation.


Circular Motion 

  • An object is said to be in a circular motion when it rotates around a center, in a circle of radius ‘r’.The wheels on the bicycle are in a circular motionThe wheels on the bicycle are in a circular motion
  • Distance covered by the object in one round = circumference of the circle = 2πr
  • Speed = Distance / time = 2πr / t
  • However, the displacement in a circular motion is zero as the object returns at its initial or reference point after one rotation.

Try yourself:A particle is moving in a circular path of radius r.
Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9
The displacement  after half a circle would be:
View Solution

Summary

  • Distance: The total path length covered by the object from its reference point. 
  • Displacement: The shortest distance covered by an object from its reference point. Displacement has both magnitude and direction.
  • Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9
    Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9
  • Equations of motions
    (i) v = u + at
    (ii) s = ut + 1 / 2at2
    (iii) 2as = v2 – u2

Try yourself:Which of the following can sometimes be ‘zero’ for a moving body?

i. Average velocity

ii. Distance travelled

iii. Average speed                           

iv. Displacement

View Solution

The document Chapter 8 - Motion Notes | Study Science - Short Notes Class IX - Class 9 is a part of the Class 9 Course Science - Short Notes Class IX.
All you need of Class 9 at this link: Class 9

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