Devendra motion

MOTION

Questions:-

   12/8/2016

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

2-8-2016

2) GRAPH

    

Graph of the above table:-

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