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Motion

Motion relates to CBSE/Class 9/Science/Unit 3-Motion, Force and Work

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Motion Lessons

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Answered on 18/04/2024 Learn CBSE/Class 9/Science/Unit 3-Motion, Force and Work/Motion

Nazia Khanum

Distance-Time Graph for Uniform and Non-Uniform Motion Uniform Motion: In uniform motion, the object covers equal distances in equal intervals of time. The distance-time graph for uniform motion is a straight line inclined to the time axis. Non-Uniform Motion: In non-uniform motion, the object covers... read more

Distance-Time Graph for Uniform and Non-Uniform Motion

Uniform Motion:

  • In uniform motion, the object covers equal distances in equal intervals of time.
  • The distance-time graph for uniform motion is a straight line inclined to the time axis.

Non-Uniform Motion:

  • In non-uniform motion, the object covers unequal distances in equal intervals of time.
  • The distance-time graph for non-uniform motion is curved.

Solution:

Given Data:

  • Initial velocity (u) = 0 m/s (as the bus starts from rest)
  • Acceleration (a) = 0.1 m/s²
  • Time (t) = 2 minutes = 120 seconds

(a) Speed Acquired:

  • Using the equation of motion: v=u+atv=u+at
  • v=0+(0.1×120)v=0+(0.1×120)
  • v=12 m/sv=12m/s

(b) Distance Travelled:

  • Using the equation of motion: s=ut+12at2s=ut+21at2
  • s=(0×120)+12(0.1×1202)s=(0×120)+21(0.1×1202)
  • s=0+12(0.1×14400)s=0+21(0.1×14400)
  • s=12(1440)s=21(1440)
  • s=720 ms=720m

Summary:

  • The speed acquired by the bus is 12 m/s12m/s.
  • The distance travelled by the bus is 720 m720m.
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Answered on 18/04/2024 Learn CBSE/Class 9/Science/Unit 3-Motion, Force and Work/Motion

Nazia Khanum

Uniform Acceleration Definition: Uniform acceleration refers to a situation where an object's velocity changes at a constant rate over time. In other words, the object's speed increases or decreases by the same amount in each successive equal interval of time. Acceleration of a Body with Uniform Velocity:... read more

Uniform Acceleration

Definition: Uniform acceleration refers to a situation where an object's velocity changes at a constant rate over time. In other words, the object's speed increases or decreases by the same amount in each successive equal interval of time.

Acceleration of a Body with Uniform Velocity: When a body is moving with uniform velocity, its acceleration is zero. This means that the object maintains a constant speed and direction, hence no change in velocity, and consequently, no acceleration.

Magnitude of Displacement for a Particle Moving Over Three Quarters of a Circle

Given:

  • Particle moves over three quarters of a circle of radius rr.

Calculation:

  1. Circumference of the Circle:

    • Circumference CC of a circle with radius rr is given by C=2πrC=2πr.
  2. Three Quarters of the Circle:

    • Three quarters of the circumference is 34×2πr43×2πr.
  3. Magnitude of Displacement:

    • The displacement is the shortest distance between the initial and final positions.
    • When a particle moves over three quarters of a circle, its displacement is equal to the diameter of the circle.
    • Diameter DD of the circle with radius rr is given by D=2rD=2r.

Result:

  • The magnitude of the displacement for a particle moving over three quarters of a circle of radius rr is equal to 2r2r.
 
 
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Answered on 18/04/2024 Learn CBSE/Class 9/Science/Unit 3-Motion, Force and Work/Motion

Nazia Khanum

Solution: Given Data: Initial velocity, u=54 km/hu=54km/h Final velocity, v=72 km/hv=72km/h Time, t=10 secondst=10seconds Conversion: To perform calculations, we need to convert velocities from km/h to m/s. Conversion: 1 km/h = 13.63.61 m/s Converting Initial Velocity: u=54 km/h×13.6=15 m/su=54km/h×3.61=15m/s Converting... read more

Solution:

Given Data:

  • Initial velocity, u=54 km/hu=54km/h
  • Final velocity, v=72 km/hv=72km/h
  • Time, t=10 secondst=10seconds

Conversion: To perform calculations, we need to convert velocities from km/h to m/s.

Conversion: 1 km/h = 13.63.61 m/s

Converting Initial Velocity: u=54 km/h×13.6=15 m/su=54km/h×3.61=15m/s

Converting Final Velocity: v=72 km/h×13.6=20 m/sv=72km/h×3.61=20m/s

(i) Acceleration (aa):

Formula: a=v−uta=tv−u

Substituting Values: a=20 m/s−15 m/s10 sa=10s20m/s−15m/s

Calculation: a=5 m/s10 s=0.5 m/s2a=10s5m/s=0.5m/s2

(ii) Distance Covered (ss):

Formula: s=ut+12at2s=ut+21at2

Substituting Values: s=(15 m/s×10 s)+12×0.5 m/s2×(10 s)2s=(15m/s×10s)+21×0.5m/s2×(10s)2

Calculation: s=(150 m)+0.5×5×100=150+250=400 ms=(150m)+0.5×5×100=150+250=400m

Answer: (i) Acceleration a=0.5 m/s2a=0.5m/s2 (ii) Distance Covered s=400 ms=400m

Therefore, the bus accelerates at 0.5 m/s20.5m/s2 and covers a distance of 400 m400m during this interval.

 
 
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Answered on 18/04/2024 Learn CBSE/Class 9/Science/Unit 3-Motion, Force and Work/Motion

Nazia Khanum

Given Information: Speed of the car: First half hour: 30 km/h Second hour: 25 km/h Third hour: 40 km/h Step 1: Calculate the total distance traveled Distance covered in the first half hour: 30 km/h×0.5 h=15 km30km/h×0.5h=15km Distance covered in the second hour:... read more

Given Information:

  • Speed of the car:
    • First half hour: 30 km/h
    • Second hour: 25 km/h
    • Third hour: 40 km/h

Step 1: Calculate the total distance traveled

  • Distance covered in the first half hour: 30 km/h×0.5 h=15 km30km/h×0.5h=15km
  • Distance covered in the second hour: 25 km/h×1 h=25 km25km/h×1h=25km
  • Distance covered in the third hour: 40 km/h×2 h=80 km40km/h×2h=80km

Total distance = 15 km+25 km+80 km=120 km15km+25km+80km=120km

Step 2: Calculate the total time taken

Total time taken = 0.5 h+1 h+2 h=3.5 h0.5h+1h+2h=3.5h

Step 3: Calculate the average speed

Average speed = Total distance / Total time taken

Average speed = 120 km/3.5 h=34.29 km/h120km/3.5h=34.29km/h

Step 4: Final Answer

Therefore, the average speed of the car is 34.29 km/h34.29km/h.

 
 
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