AARUSH SINHA Roll No.-20 Chapter-3 MOTION

CHAPTER-3

MOTION

Questions:-

21-JUNE-2016

1) Define the scalar quantity and name 5 physical quantities which are scalar and give reasons why these physical quantities are scalar?
ANS- Scalar quantity is the physical quantity in which only magnitude is measured and                       direction is not specified .
          Eg- Time, Speed, Volume, Temperature, Area.
                  All these quantities are scalar quantities because there direction is not specified, their only                                 magnitude is measured

24-JUNE-2016

2) GRAPH
    



Graph of the above table:-

It is a non-uniform motion.




     
EXAMPLES

(Both theoretical and numerical questions) 


1. An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the                object?
ANS- Total distance covered by the object = 16+16 = 32m
           Total time taken = 4+2 = 6s
           Therefore, average speed of the object =    32  = 5.33 m s-1
6

2 . The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the      trip took 8 h, calculate the average speed of the car in km h–1 and m s–1.
ANS- Total distance covered by the object = 2400-2000 = 400km
           Total time taken = 8h
           Therefore, average speed of the object =   400 = 50 km h-1
8
           In  m s–1  = 13.9 m s–1.

3 . Usha swims in a 90 m long pool. She covers 180 m in one minute by swimming from one end to         the other and back along the same straight path. Find the average speed and average velocity of           Usha. 
ANS- Total distance covered by Usha in 1 min is 180 m.
           Displacement of Usha in 1 min = 0 m
           Average speed of the object =    180 = 3 m s-1
                                                                                     60
             
          Average velocity = Displacement   =  0m = 0 m s-1
                                        Total Time          60

4.Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m s–1 in 30 s.    Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s.      Calculate the acceleration of the bicycle in both the cases.
ANS- In the first case 
          U=0 , V=6 m s-1 , t = 30sec , a = ?
          a = V-U   =  6-0 =  1  m s-2  =  0.2 m s–2
                   t           30     5 

In the second case 
          U=6 m s-1 , V=4 m s-1 , t = 5sec , a = ?
          a = V-U   =  4-6 =  -2  m s-2  =  -0.4 m s–2
                   t             5       5 

5. A train starting from rest attains a velocity of 72 km h–1 in 5 minutes. Assuming that the                     acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for               attaining this velocity.
ANS- U=0, V= 72 km h-1 = 20 m s-1, t = 5min = 300sec
     (i) a = V-U   =  20-0 =  1  m s-2 
                   t           300    15  
     (ii) by IInd equation of motion
           S= ut + 1/2 at*2
           S= 0X300 + 1/2 X 1/15 X 90000
           S= 3000m = 3km

6. A car accelerates uniformly from 18 km h–1 to 36 km h–1 in 5 s. Calculate 
(i) the acceleration and 
(ii) the distance covered by the car in that time.
ANS- U= 18 km h-1 = 5 m s-1 , V= 36 km h-1 = 10 m s-1 , t = 5 s  
     (i) a = V-U   =  10-5 =  1  m s-2 
                   t            5          
     (ii) by II nd equation of motion
           S= ut + 1/2 at*2
           S= 5X5 + 1/2 X 1 X 25
           S= 25 + 12.5 = 37.5 m

7. The brakes applied to a car produce an acceleration of 6 m s-2 in the opposite direction to the             motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels         during this time.
ANS- a = -6 m s-2 , t = 2 sec , V= 0 m s-2 
          by I st equation of motion 
          V=U+at
          0= U+(-12)
          0=U-12
          U= 12 m s-1
           
         by II nd equation of motion
         S= ut + 1/2 at*2
         S= 12X2 + 1/2 X -6 X 4
         S= 24 - 12 = 12 m



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