Motion 8 lesson (9th class)

knowledgeworld.blogger .com

Lesson no. - 8 

Lesson name- Motion

Text book- NCERT 

  Motion 8 lesson Ncert solution

{Part 1}

 Q]What is the motion? & explain with example.

Ans:    When the object change its position from one place to another is called as motion.

   Ex -

• A moving car
•  ball 
• cycle 
• pendulum

Page no.100 

  Q.1] An object has moved through a distance can it have zero displacement if yes support your answer with an example.

Ans:   Yes because when an object move and come on it initial position it means it's displacement is zero.

Ex -

Q.2 ]   A farmer moves along the boundary of a square field of side 10 metre in 40 seconds what will be the magnitude of displacement of farmer at the end of 2 minutes 20 seconds for its initial position?

Ans:

case.1.( speed of farmer)- {Sf}

Sf=distance/ time

Sf=40/40 (m/s)
✓Sf =1 m/s

 Case.2.

S=1m/s (metre per second )
Time = 2 minute 20 seconds
1min=60sec
2min=2×60
          =120
t (Time)= 120×2
              =140sec

                                 
d (distance )  =speed (s)×Time(t) 

=1×140
=140m/s

= 140/40
=7/2
=3.5

                :. At the end of 2 minutes 20 seconds from his initial position it cover 3.5 metre.

Displacement=AC
     2      2     2
AC =AB +BC
           2      2
     =10+ 10
=100+100
=√200
     

Q.3) Which of the following is true for displacement.

1] It can not be zero.
  -------False

2] Its magnitude is greater than the distance travelled by the object.

 ------- false

Page  no.102

Q 1] Distinguish between speed and velocity.
Ans: 
Speed---

• Speed does not have a direction.
• Speed is measured in metre per second
• Speed it is the  sector quantity.
• Speed can never be negative or zero ,it is only positive.
• Speed never decrease with time.
• Speed main aur may not be equal to velocity.
• v=d/t, were v- (s)speed, d- distance & t- time

Velocity---

• Velocity is speed with direction.
• Velocity is aslo measured  in metre per second
• Velocity is vector quantity
• Velocity can be zero negative or positive.
• Velocity can decrease with time.
• an object may possess different velocity but the same time. 
• ∆x/∆T, were. ∆T= Time of arrival & ∆x =displacement.

Q.2] Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Ans:   Average velocity would be equal to average speed when the total distance travel = the net displacement of a particle.This happen when a particle moves along a straight line in a fixed direction.

Q.3]. What does the odometer of an automobile measure?
Ans:
Odometer is the device which measure the distance travelled by an automobile based on ,the perimeter of the wheel as it  rotates.

Q.4]. What does the path of an object look like when it is in uniform motion?
Ans:
Uniform motion is the motion in which the velocity vector remain constant at every point of time so, that the path of an object look like straight line when it is in uniform motion &  object is moving in a straight path with uniform acceleration.

Q.5]  During an experiment, a signal from a spaceship reaches the ground station in 5 what was the distance of the spaceship from the ground .What was the distance of the spaceship from the ground station the signal travels at the speed of light that is 3×10 power of 8 m/s.

Ans:  1 min=60 sec
          5min= 5× 60=300sec
So, distance= speed* time
                      =300*3*10power of 8
                      =900×10 to the power of 8
                      =9×10  to the power of 10

Page no. 103

Q.1] When will you say a body is in

(i) uniform acceleration?
Ans: When an object change its velocity by equal interval of time then we say that the object is in uniform acceleration.

(ii)Non uniform acceleration.                     
Ans:.      An object change it velocity by unequal interval in equal interval of time then we say that object is in non uniform acceleration .                                                       

                                                                              

Q.2] A bus decreases its speed from 80km h to 60 km h in 5 s . Find the acceleration of the bus.

 Ans: initial velocity u= 80 km h.                                    Final velocity v= 60km h.                        

Time t = 5 s.                 

SI  unit of velocity= m/s         

                                  

                    1km= 1000m.          

u= 80 *1000= 80000m/s

v= 60*1000=60000 m/s

Acceleration of the bus= v-u / t= 60,000-80,000/5=

-20,000/5=

-4000m/s

:.Therefore the acceleration of the bus is -4000 metre per square.

Q.3] A train starting from a railway station and moving with uniform acceleration attains a speed 40 kilometre in 10 minutes. Find its acceleration.

Ans: u=0( initial position)

         v= 40 (final position)

         t= 10 min

Acc= ?

Formula of acceleration= v-u/ t

                                           =40-0/10

                                             =40/10=4km per min

             

Exercise

Q.1) An athlete completes one round of a circular track of diameter 20 metre in 40 seconds what will be the distance covered and displacement at the end of 2 minutes 20 seconds?

Ans:. the diameter of circular track =200
 radius of circular track is equals to 200/2 =100

• Time taken by the athlete for one round =40 sec
• Distance covered by athlete is in one round. (s) =2×πr    
• 2×22/7×100 
• Speed of athlete =distance /time
• 2×2200/7×40=4400.                                                                distance= 2200m                        Displacement=200m

Q.2 ] Joseph jogs  from one end  to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turn around and job 100m back to point C in another 1 minute .What are Joseph 's average speed and velocity in jogging from  to( a) from A to B & (b) from A to C?

Ans: 

Total distance  covered from AB=300m
Total time taken =2*60 +30s =150s

Therefore, average speed from AB=distance divided by total time=300/150=2m/s

Therefore, velocity from AB= displacement AB/time= 300/150=2m/s

Total distance covered from AC=AB+BC
                                                        =  300+200
                                                        =  500
Total time taken from A to C= time taken for AB&BC
=(2*60+30)60s
=210s
:. Therefore, average speed from AC=Total distance/total time=400/210= m/s
                                                   =1.904m/s
Displacement(s) from A to C=AB-BC
                                                  =300-100m
                                                  =  200m
:.  Therefore  velocity from AC=displacement/time=200/210m/s=0.952m/s

      Q. 3]   Abdul while driving to school computes the average speed for his trip to be 20 km per hour on his return trip along the same route there is less traffic and average speed is 30 kilometre per hour what is the average speed of Abdul trip ?

Sol'n :   

Average speed=20km/h

Due to less traffic average speed is=30km/h

Average speed =?

(By formula 2xy/x+y)=2*20*30/20+30

                                     =1200/50

                                     =24km/h