ADITI-Roll No.45 (3)

        CHAPTER - 3 MOTION 

Date : 21.06.16

Question: Define the scalar quantity and name 5 physical quantities which are scalar .

Answer : Scalar quantity is the physical quantity in which only magnitude is measured and direction is not specified . For eg : speed , distance,time.
 These all quantities are scaler because there direction is not specified only their magnitude is measured .
                                                             Date : 24 .06.16

Graph :
                                             

                                                            Graph of the above table

                                                   

                                                                   It is not a uniform motion 

                                    Examples 

                     Example 1 :

    Question : An object travels 16 m in 4 s and then another 16 m in 2 s . What is the average                         speed  of  the object  ?

                  Answers : The total distance covered by the object  = 16 + 16 = 32 m 

                                   Total time taken by the object = 4 s + 2 s = 6 s                        

                                    average speed = Total distance travelled  =  32 m = 5.33 m/s

                                                                    Total time taken               6 s 

                                                           

                                       So , the average speed of the object is 5.33 m/s 

    Question : Usha swims in 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 .  

     Answer : The total distance covered by  Usha in 1 minute = 90 + 90 = 180 

                             displacement of Usha in 1 min = 0 m 

                               Average speed = Total distance covered     

                                                                 Total time taken    

                                                       =  180 m      =   180 m     x   1 min 

                                                              1 min            1 min          60 secs

                                                         = 3 m/s 

                                Average velocity =  Displacement 

                                                                  Total time taken 

               

                                                     =  0 m 

                                                         60 s       = 0 m/s 

                                        

                       so , we find that the average speed of Usha is 3 m/s and her average 

                                              velocity is 0 m/s .

 

                                THEORETICAL 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

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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

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           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