Tuesday, 1 September 2020

Numericals of Ex-2(A) (Q1 to Q7 )of chapter 2 based on motion in One dimension of class-9

 1. The speed of a car is 72 kilometre per hour .Express it in metre per second.

Ans. Speed of car=72km/h

            Speed of a car=72kmh-¹

                       =72×1km / 1h   (1km=1000m, 1h=60minutes, 1minute=60 seconds, so 1h=60×60s)

                       =72×1000m / 60×60 s

                       =72 × 5m /18 s

                       =4×5m / 1s

                       =20 ms-¹.

2. Express 15 metre per second in kilometre per hour.

Ans. Speed=15m/1s

Speed=15× 1m/s (1km/h =5/18 ms-¹ , 18/5 ×1km/h =1ms-¹)

=15 ×18/5 kmh-¹

=3×18 kmh-¹

=54 kmh-¹.

Express each of the following in metre per second: a)1kmh-¹ b)18kmh-¹

Ans. 1km/h = 1000m/ 60×60s  (1km=1000m)

         =10m/36s                            (1h=60×60s)

         =5m/18s

         =0.278 ms-¹.

18 km/minute =18×1000m/60s   (1 minute=60s)

                           =18×100m/6s

                           =3×100m/s

                           =300m/s.

4. Arrange the following speeds in increasing order: 10ms-¹,1km min-¹,18km h-¹.

Ans. Convert them  to their Si units and then compare their speeds & arrange them from smaller to bigger.

1m/s=10ms-¹

1km/min = 1000m/60s

                  =100m/6s

                  =16.666m/s

18km/h =18×1000m/60×60s

              =18×5m/18s

              =1m/s.

1ms-¹ < 10ms-¹ < 16.67ms-¹

18kmh-¹ < 10ms-¹ < 1kmmin-¹.

5. A train takes 3 hour to travel from Agra to Delhi with a uniform speed of 65 kilometre per hour. Find the the distance between the two cities?

Ans. Speed=65kmh-¹

Time=3h

Distance=?

Speed= distance/ time. 

Speed× time= distance

Distance= speed × time

                =65km/h × 3h

                =65×3 km

                =195 km.

6. A car travels first 30 kilometre with a uniform speed of 60 kilometre per hour and then next 30 km with a uniform speed of 40 kilometre per hour .calculate: a) the total time of journey b) the average speed of the car

Ans. Whenever  we are asked to find average speed then we have to find Total distance travelled and the total time taken for the  the whole journey .

d1 = 30km

S1 = 60 kmh-¹

T1 = ?

Speed= distance/ time

Speed= distance/ time

Speed× time= distance

Time= distance/ speed

T1 = d1 / s1

     = 30km/60kmh-¹

    =1/2 h

   =0.5 h

d2 = 30km

S2 = 40km/h

T2 =?

T2=d2 / s2.

     =30km / 40kmh-¹

    =3/4 h

Total time of journey=T1 + T2

                          =0.5 h + 0.75 h

                          =1.25 h

 =1h + 0.25 h

=1h + (25/100)  ×1 h  (1h=60 minutes)

=60 min + (1/4)× 60 min

=60 min + 15 min

=75 minutes.

Average speed = Total distance travelled/ total time taken.

 Average Speed= (d1 + d2 ) / (T1 + T2)

                            =(30km + 30 km)/(0.5 h + 0.75h)

                            = 60 km /1.25 h

                            = 48 km/h

7. A train takes 2 hour to reach station be from station, and then three hours to return from station B to station A. Distance between the two station stage 200 kilometre. Find: a) the average speed b) the average velocity of the train.

Ans. Total distance travelled= B to A + A to B

=200 km + 200km

=400km

Total time taken=A to B + B to A

=2h + 3 h

= 5h.

Average speed= Total distance travelled/ total time taken 

average speed= 400 km / 5h

                           =80 kmh-¹

The average velocity of the train  is zero because starts from A to B and then returns back from B to A . Displacement is zero.

Average velocity= total displacement/ total time taken.

Average velocity= 0km / 5h

                              = 0 kmh-¹

Displacement is zero.so average velocity would also be zero.


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