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Illustrations

Example-1

A particle moves along a circle of radius R. When it has covered quarter of the circumference, its displacement is ________.

  1. R2 
  2. Description: 46539.png 
  3. Description: 52428.png 
  4. zero
Solution

Ans (3)

 

Description: 52492.png


Displacement of the particle is given by

 

Description: 46555.png

 

Description: 46569.png

 

Description: 46577.png

 

(By Pythagoras theorem AB2 = OA2 + OB2)
 

 

Example-2

Light takes 8.5 min to travel from the sun of the earth. The velocity of light 3 × 105 km s−1. Then the distance of the sun from the earth is ________.

  1. 1.53 × 108 km
  2. 3 × 108 m
  3. 4 × 108 km
  4. 2.55 × 106 km
Solution

Ans (1)

 

Distance of sun from earth

 

= Speed of light × Time taken by light

 

= 3 × 105 km s−1 × (8.5 × 60) s

 

= 1.53 × 108 km
 

 

Example-3

The distance travelled by a car in the first 5 s as per the given velocity–time graph is ________.

  1. 100 m
  2. 25 m
  3. 4 m
  4. 0 m
Solution

Ans (1)

 

Description: 52510.png

 

Displacement in 5 s

 

= Area of rectangle OABC

 

OA × OC = 5 × 20 = 100 m
 

 

Example-4

A car acquires a velocity of 72 km h−1 in 10 s starting from rest. Calculate

  1. the acceleration
  2. the average velocity
  3. the distance travelled in this time.
Solution

Given: Initial velocity u = 0 m s–1

 

Final velocity Description: 46637.png

 

Time taken t = 10 s

  1. Acceleration Description: 46650.png
     
    Description: 46659.png
     
    Acceleration, a = 2 m s–2
  2. Average velocity Description: 46720.png
     
    Description: 46727.png
  3. Distance travelled s = Average velocity × Time
     
    = (10 m s–1) × (10 s) = 100 m
     
    Alternatively Description: 46736.png
     
    = 0 m s–1 × 10 s + Description: 46746.png m s–2 × 10 s × 10 s
     
    = 0 m + 100 m = 100 m

 

Example-5

A stone thrown vertically upwards takes 3 s to attain the maximum height. Find the

  1. initial velocity of the stone
  2. maximum height attained by the stone (g = 9.8 m s–2)
Solution

Given: Acceleration due to gravity, g = 9.8 m s–2

 

Final velocity, v = 0 m s–1

 

Time, t = 3 s

 

Distance = Height = ?

 

Here, u is upwards and g is downwards. We take u as +ve and g as −ve.

  1. To find the initial velocity
     
    v = u + at (here a = −g)
     
    v = u − gt
     
    u = v + gt
     
    = 0 m s–1 + 9.8 m s–2 × 3 s
     
    = 29.4 m s–1
  2. To find the height
     
    s = ut Description: 46787.png
     
    = (29.4 m s–1 × 3 s) – Description: 46798.png (9.8 m s–2 × (3 s)2
     
    = 88.2 m – 44.1 m = 44.1 m
     
    Therefore, the maximum height is 44.1 m.




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