The Motion, Class 9th Notes

THESE NOTES ARE PREPARED ACCORDING TO JKBOSE SYLLABI IN SIMPLIFIED AND CONCISE  PATTERN

Q. Define Motion?

Ans:  Motion is change in position of a body with respect to time.

Motion is a relative quantity and is of two types,
Uniform motion.
Non Uniform motion.

Q. What is difference between Distance and Displacement?

Ans:

Distance is the actual length of path covered by moving body.
Displacement is shortest  distance between initial and final position of body.

Distance is scaler quantity, Displacement is vector quantity

Distance cannot be zero, Displacement can be zero.

Q. What is uniform and Non uniform motion?

Ans:
Uniform motion:  A body is said to be in uniform motion if it travels equall distance in equall intervals of time.
The distance time graph of Uniform motion is straight line.

Non Uniform motion:  The body is said to be in non uniform motion if it travels unequal distance in equall intervals of time.

The distance time graph is curved line.

Q. Define Speed? Give its SI unit.

Ans: Speed is the distance travelled by the body per unit time.


It is denoted by "V" and is given by,      
                              Distance   
                                Time

   i.e,    V  =      S/T

The SI unit  of Speed is Meters/Second.
Speed is Scaler quantity.

Q. What is Velocity?

Ans: Velocity is distance travelled by body per unit time in a given direction.
It is also defined as Displacement produced per unit time.

It is given by,

Velocity    =           Distance  
                                  Time

The SI Unit of Velocity is Meters/Second. It is a vector quantity.

Q. What is Acceleration? What are its types.

                                     Or

Q. When will you say a body is in (i) uniform acceleration? (ii) nonuniform acceleration?

Ans: Acceleration is defined as rate of change in velocity per unit time.

It is denoted by "A", and is given by,

Acceleration =        Change in Velocity    
                                      Time

Acceleration    =    Final Vel -   Initial Vel
                                       Time

Its SI Unit is meters/Second square

Acceleration is of two types.

1. Uniform Acceleration: When the velocity increases by equal amounts in equal intervals of time. Eg Free fall.

2. Non Uniform Acceleration: When the velocity changes at non uniform rate.



Q. What are the three equations of motion.


Ans: There are three equations of motion for bodies which travel with uniform acceleration.

1. Velocity Time Equation: It is first equation of motion
                  V  = U  +   AT

2.   Position Time Equation: It is second equation of motion.
                                         
                                           2
            S    =  UT + 1/2 AT

3.  Position Velocity Equation: It is third equation of motion.
             2            2
           V   -    U   =   2  AS


Q. Derive the First Equation of motion.
                                 Or

Derive the relation of Velocity and Time.

Ans: The first equation of motion gives velocity time relation. i.e,

                V  = U  +   AT
Where, V is final Velocity, U is initial Velocity, A is acceleration and T is time.

Consider a body of Initial velocity U, having uniform  acceleration A, final velocity V after time T,

           A   =    V  -  U
                           T
           AT  =   V  -   U
           V   =  U + AT


Q. Derive Second equation of motion.
                           Or
Derive equation for position time relation.

Ans:  The second equation of motion givee position Time Relation. i.e,
                                       2
           S =  UT +  1/2 AT

Where, S  is distance travelled, U is initial velocity,  T is the time, A is Acceleration.

The distance "S" can be found by average velocity,

   Avg Velocity  =    Initial vel  + Final Vel    
                                             2

Distance travelled = Velocity  × Time

   S  =      U   +   V     ×   T
                     2

Acc. to Ist Equation of motion,

        V   =    U  +   AT

So,
    S    =     U  + U + AT     ×  T
                      2

    S   =     2 U  +   AT  × T
                       2
   
                                      2
       S  =      2 UT  +  AT
                           2
                                      2
       S =     UT +  I/2 AT


Q. Derive third equation of motion.
                          Or
  Derive the equation for position velocity relation.

Ans:  The third equation of motion gives Position Velocity relation.

It is given by,
                  
              V² = U² + 2AS

Where V is final velocity, U is Initial Velocity, A is Acceleration, S is distance travelled.

Third Equation of Motion 
                  

We know that; 
         V = U + AT
=>     V - U = AT
or T = (V - U)
               A
Also we know that
Distance = average velocity X Time

.: S = (V+U)   ×   (V-U)
              2              A

=>     S = (V²– U²)
                    2A

=>   2AS = V² – U²


This is the third equation of motion.

Q. What is uniform circular motion? Give its examples.

Ans: When a body moves in circular path with uniform speed, Its motion is said to he uniform circular motion.

The direction keeps changing, thus it is tyoe of accelerated motion.

The Uniform circular motion is given by,

      V    =          2 π R   
                             T

The examples of uniform circular motion are,
1. The motion of satallite around earth,
2. The motion of seconds hand in clock

3. The motion of cyclist in circular track


Important Textual Questions:

Q. During experiment, signal from spaceship reaches ground in 5 minutes. What is the distance of spaceship from ground.

Sol:
Time Taken : 5 min
  I.e, 5 ×60 = 300 sec
                                           8
Speed of signal=   3 × 10     m/s

Distance =    V  ×  T
                                            8
                =    300 × 3 × 10
                                  10
                =    9  × 10      m

Q.  A bus decreases speed from 80 km/h to 60 km/h in 5 seconds. Find its acceleration.

Sol:
     Initial speed, U  =   80 km/h
                       =    80  ×   1000
                               60  ×  60
                       =    22.22 m/s

Final  speed  V =   60 km/h  = 16.66 m/s

Time, T    =       5 sec

Acceleration, A   =         V   -   U
                                               T
                             =   16.66  -   22.22
                                            5
                                                                        2
                               =  - 5.56/5   = -1.11 m/s
Minus sign indicates deceleration. 

                   
Q.  A train starts from railway station and moves with uniform acceleration attaining the speed of 40 km/h in 10 minutes. Find the acceleration.

Sol:
    Initial speed, U = 0
    Final speed, V =  40 km/h  =  11.11 m/s
    Time taken , T = 10 min = 600 sec

Acceleration, A =   V  -    U
                                       T

                       A   =   11.11 - 0   
                                     600
                                               -2              2
                    A  =   1.85 ×  10      m/s


Q. What is nature of distance time graph for uniform and non uniform motion of object?

Ans:  In uniform motion, the graph is straight line.

In non uniform motion, the graph is curved.


Q. When the distance time graph is straight line parallel to time axis?

                                                          Or

What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Ans:    When the object is stationary.

Q. What can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis?

Ans:  When the body is moving with uniform speed and there is no acceleration.


Q. What is measured by area occupied below velocity time graph?

Ans: Area under velocity time graph gives distance travelled by body.

Q.  A bus starts from rest, moves with uniform acceleration of 0.1 m/s2 for 2 minutes. Find speed and distance.

Ans: Here,
                                     2
U  =   0,   A = 0.1 m/s,  T =  120 sec

Speed, V  =   U  +   AT
                 =   0 + 0.1 × 120
                =   12 m/s
                                                 2
Distance, S  =   UT  + 1/2 AT
                                                                     2
                    =   0 × 120 + 1/2 × 0.1 × 120

                    = 720 METERS.

Q. A Trolley while going on inclined plane, Acceleration is 2 cm/s2 from rest. What is the velocity 3 seconds after start.

Ans: Here,
Acceleration = 2 cm/s 2, Time = 3 s,
Initial Velocity = 0, Final Velocity =  V

      V   =     U   +   AT
       V   =     0  +   2× 3
      V    =      6 CM/SECOND


Q. A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If the acceleration of the stone during its motion is 10 m s–2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

 
Sol: Here,
   U =   5 m/s,     A  =  - 10 m/s,  V =  0.

We have,
  Distance::
      2       2  
    V   -   U    =     2 A S
             2        2
  S =   V    -  U
              2A

  S  =    0 - 25
            2 (-10)

  S  =    1.25 m.

Time:::
      V  =    U  +    AT
      T  =   V  -   U
                    A  
       T  =   0 -  5
                   -10
        T  =   0.5 sec.

Q. Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic and the average speed is 30 km h–1. What is the average speed for Abdul’s trip?

 

Ans:  Time taken to reach T1,
         =      Distance        =        x    
                   Speed                     20

Time taken to return, T2
  =        Distance        =        x    
             Speed                     30

Total time taken  =  T1  +  T2
      =    X/20  +   X / 30

       =      3X    +      2X 
                     60
       =        X    
                  12

Total distance = 2x,

Speed=                   2x              =  2X     ×  12
                                 X                     X
                                12 
Speed   =   24 km/h



Q. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Ans: Here,

R   =   42250 km,    T=  24 hours, Speed =  V

We have,
        V     =         2   π     R   
                                 T

     V  =   2  ×   3.14  ×  42250
                         24 
                                 4
   V   =       1.1  × 10      Km/h.

Q. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Ans: Yes, an object can have zero displacement even if it moved through a distance.

Example: If an object moves 5m from Point A to B and then 5m back to A. It moved the distance of 10 m but its displacement is Zero.

 



Q. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Ans: Given,

Side of Square field = 10 m

Time taken to move along the boundary = 40 s

Time for which farmer moves along the boundary = 2 m, 20 sec i.e. 140 s

Now,

Perimeter of field = 4 x Side = 4 x 10 = 40 m

Distance covered in 1 sec = Perimeter of field/Time taken

Distance covered in 1 sec = 40/40 i.e. 1 m

Distance covered in 140 s = Distance covered in 1 sec x 140 i.e. 140 m

i.e. Farmer makes Three and a half rounds along the field in 140 seconds (2 min, 20 sec)


Now,

 

Displacement = √(10)^2 + (10)^2

Displacement = √100 +100

Displacement = √200

Displacement = 14.14 m









Q. Which of the following is true for displacement?

(a) It cannot be zero.

(b) Its magnitude is greater than the distance travelled by the object.

Ans: None of the Above.

Displacement may be less than or equal to the distance. It can be zero as well.





Q. Distinguish between speed and velocity.

Ans: 
 

Speed	Velocity
It is distance travelled per unit time	It is distance travelled per unit time in a particular direction
It is Scaler quantity	It is Vector quantity
Speed = Distance/Time	Velocity = Displacement/Time
It can be positive or zero	It can be positive, negative or zero

 

 

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

Ans: We know that;

Average Speed = Distance/ Time &

Average Velocity = Displacement/ Time

Therefore, when the distance and the displacement of an object is same, the average speed will be equal to average velocity.

 

Q. What does the odometer of an automobile measure?

Ans: The odometer of an automobile measures the distance travelled by the object.

 

Q. What does the path of an object look like when it is in uniform motion?

Ans: The path of an object in uniform motion can be straight line, curved line or in circle.

However path of an object in the uniform motion on a distance time graph will be straight line at an angle with the X axis.

Q. A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of – 0.5 m/s2. Find how far the train will go before it is brought to rest.



Ans: Given,

Speed of the train = 90 km/h i.e. 90 x 5/18 = 25 m/s

Acceleration = -0.5 m/s^2 (Minus sign indicates that the train is decelerating)

Distance at which the train is bought to rest =?

We Know,

V^2 – U^2 = 2AS

0 – (25)^2 = 2 x -0.5 x S

0 – 625 = -1S

S = -625/-1

S = 625 m

Therefore the train will be bought to rest at 625 m.

Q. A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start?


Ans: Given,

Initial Velocity = 0 m/s

Final Velocity =?

Acceleration = 4 m/s^2

Time = 10 s

Distance = ?

We Know,

S = UT + ½ AT^2

S = 0 x 10 + ½ x 4 x (10)^2

S = 2 x 100

S = 200 m

Therefore, the car will travel 200 m in 10 seconds at a uniform acceleration of 4m/s^2

 Q. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?


Ans: Given,

Diameter of Track = 200 m

Time taken to complete one round = 40s

Total time travelled = 2 min, 20 sec i.e. 140 sec

Now,

Total rounds made in 140 seconds = 140/40 i.e. 3.5 rounds

Distance travelled in 140 sec = Rounds made in 140 seconds x Perimeter

Distance travelled in 140 sec = 3.5 x πD

Distance travelled in 140 sec = 3.5 x 22/7 x 200 i.e. 2200 m


Now, Displacement of Athlete = Diameter of track i.e. 200 m.

This is because, after 3.5 rounds around a circular track, the athlete will be opposite to the starting point.

Q. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Ans:

Distance covered from A to B = 300 m

Time taken to Jog from A to B = 2 m, 30 s i.e. 150 s

Therefore, Average Speed = Distance/Time

Average Speed = 300/150 = 2 m/s

& Average Velocity = Displacement/Time

Average Velocity = 300/150 = 2m/s

Distance covered from A to C = 400 m

Time taken to Jog from A to C = 210 s

Therefore, Average speed = Distance/Time

Average Speed = 400/210 = 1.9 m/s

& Average Velocity = Displacement/Time

Average Velocity = 200/210 = 0.95 m/s

Q. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s2 for 8.0 s. How far does the boat travel during this time?

Ans: Given,

Initial Velocity (U) = 0 m/s

Acceleration = 3 m/s^2

Time taken = 8 s

Distance travelled by the boat =?

We Know that,

S = UT + ½ AT^2

S = 0 x 8 + ½ x3 x (8)^2

S = ½ x 3 x 64

S = 96 m

Therefore, the boat travels 96 m in 8 seconds.





Q. Fig. shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest?

Ans: Since the angle of object B is more than other two, thus it is travelling fastest.


(b) Are all three ever at the same point on the road?

Ans: For all the three objects to be on same point on the road, their lines should coincide at a same point on the graph.

Therefore, all the three are not at the same point on the road.


(c) How far has C travelled when B passes A?

Ans: When the B passes A, the C has travelled 9 kilometers approximately.


(d) How far has B travelled by the time it passes C.

Ans: The B has travelled 5 kilometers approximately when it passes C.




Q. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m/s2, with what velocity will it strike the ground? After what time will it strike the ground?

Ans: Given,

Height = 20 m

Acceleration = 10m/s^2

Initial Velocity = 0 m/s

We Know,

V^2 – U^2 = 2AS

V^2 – (0)^2 = 2 x 10 x 20

V^2 = 400

V = √400

V = 20 m/s

Therefore, the final velocity is 20 m/s



Now, To know the Time after which it will hit the ground, we know:

V = U + AT

20 = 0 + 10T

T = 20/10

T = 2 Seconds




Q. The speed-time graph for a car is shown is Fig.

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

Ans:   Red coloured area represents the distance travelled by the car in first 4 seconds and Green coloured area represents the uniform motion by the car.

Q. State which of the following situations are possible and give an example for each of these:

(a) an object with a constant acceleration but with zero velocity

Ans: An object can have constant acceleration but zero velocity.

Example: Let's say a ball is thrown in vertically up with starting velocity u. Its velocity gradually drops until it reaches its maximum height, at which point it will be zero. However, because of gravity, it experiences a continuous acceleration of 9.8 m/s2.



(b) an object moving with an acceleration but with uniform speed.

Ans: An object can move with an acceleration but with uniform speed.

Example: An object travelling in a circle at a constant speed.



(c) an object moving in a certain direction with an acceleration in the perpendicular direction.

Ans: An object can move in a certain direction with an acceleration in the perpendicular direction.

Example: When one end of a string is tied around a stone, and the other used to spin the stone circle. Its acceleration is directed towards the middle in a perpendicular direction.

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 The blogger works as teacher with J&K Department of School Education.

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