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Published on: 28/09/2019
Motion
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1.
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?
2.
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?
3.
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?
4.
Establish the relation \(v^{ 2 }-u^{ 2 }=2as\) , where 'u' is the initial velocity, 'v' the final velocity 'a' the uniform acceleration and 's' is the distance covered by the body.
5.
Deduce the relation:s = ut + \(\frac { 1 }{ 2 } at^{ 2 }\) where the symbols used have their usual meanings
6.
Define acceleration .Is it a scalar or a vector quantity?
7.
Define the term average speed.
8.
Give two differences between distance and displacement
9.
Give some points of differences between distance and displacement
10.
Describe the various types of motions observed in bodies.
11.
A scotter starts from rest moves in a straight line with a constant acceleration and covers a distance of 64 m in 4s.
(i) Calculate its acceleration and its final velocity.
(ii) At what time the scooter had covered half the total distance?
12.
On a 100 km track , a train travels the first 30 km at a uniform speed of 30 km \(h^{ -1 }\)How fast must the train travel the next 70 km so as to average 40 km \(h^{ -1 }\) for the entire trip?
1.
Let one-way distance = x km
Time is taken in forwarding trip at a speed of 20 km/h = \(\frac { Distance }{ Speed } =\frac { X }{ 20 } h\)
Time is taken in return trip at a speed of 30 km/h = \(\frac { X }{ 30 } h\)
Total time for the whole trip = \(\frac { Distance }{ Speed } =\frac { x }{ 20 } +\frac { x }{ 30 } =\frac { 3x+2x }{ 60 } =\frac { 5x }{ 60 } h\)
Total distance covered = x + x = 2 x km
Average speed =\(\frac { Total\quad Distance }{ Total\quad time } =\frac { 2x }{ 5x/60 } +\frac { 2xX60 }{ 5x/60 } \)
= 24 km \(h^{ -1 }\)
2.
(a) For motion from A to B:
Distance covered = 300 m
Displacement covered = 300 m
Time taken = 2 minutes 30 seconds =2 x 60 + 30 = 150 s
Average speed = \(\frac { Distance\quad covered }{ Time\quad taken } =\frac { 300m }{ 150\quad s } =2\ ms^{ -1 }\)
Average velocity = \(\frac { Displacement }{ Time\quad taken } =\frac { 300m }{ 150\quad s } =2\ ms^{ -1 }\)
(b) For motion from A to B C:
Distance covered = 300 + 100 = 450 m
Displacement covered = AB - CB = 300 - 100 = 200 m
Time taken = 2.50 + 1.00 = 3.50 min = 210 s
Average speed=\(\frac { Displacement\quad covered }{ Time\quad taken } =\frac { 400m }{ 210\quad s } =1.90\ ms^{ -1 }\)
Average velocity= \(\frac { Displacement\quad covered }{ Time\quad taken } =\frac { 200m }{ 210\quad s } =0.952\ ms^{ -1 }\)
3.
Time taken = 2 min 20 s = 2 x 60 + 20 = 140 s
Radius,r = 100 m
In 140 s, the athlete will complete three and a half round
Distance covered = \(2\pi \) x 3.5
= 2 x \(\frac { 22 }{ 7 } \) x 100 x 3.5 = 2200 m.
At the end of his motion, the athlete will be in the diametrically opposite position
\(\therefore \) Displacement = diameter = 200 m
4.
Third equation of motion. Let a body start with initial velocity u and after covering distance s under uniform acceleration a, its velocity v in t seconds.Then
Average velocity = \(\frac { u+v }{ 2 } \)
So the distance covered in time t is given by
s = Average velocity xTime = \(\frac { u+v }{ 2 } Xt\)
v + u = \(\frac { 2s }{ t } \)
Using the first equation of motion : v = u+at or v - u = at
Multiplying equations (i) and (ii) ,we get
(v+u) (v-u) = \(\frac { 2s }{ t } \) x at or
\(v^{ 2 }-u^{ 2 }=2as\)
5.
The second equation of motion. Suppose a body starts with initial velocity u and due to uniform acceleration a, its final velocity becomes v after t. Then
Average velocity = \(\frac { Intial\quad velocity+Final\quad velocity }{ 2 } =\frac { u+v }{ 2 } \)
So, the distance covered by the body in time t is
s = Average velocity x Time
=\(\frac { u+v }{ 2 } \) x t=\(=\frac { u+(u+at) }{ 2 } Xt\) \([\therefore v=u+at]\)
\(=\frac { 2ut+at^{ 2 } }{ 2 } \)
or \(s=ut+\frac { 1 }{ 2 } at^{ 2 }\)
6.
Acceleration. In non-uniform motion, the velocity of a body changes with time It has different velocities at different instants of time and at different points of its path In such a situation we define a physical quantity called acceleration which is a measure of the change in the velocity of a body per unit time.
Acceleration is defined as the rate of change of velocity.If the velocity of a body changes from u to v in time t, then
Acceleration=\(\frac { Change\quad in\quad velocity }{ Time\quad taken } =\frac { Final\quad velocity-Initial\quad velocity }{ Time\quad taken } \)
or \(a=\frac { v-u }{ t } \)
As acceleration has both magnitude and direction, it is a vector quantity.
7.
Average speed. When the speed of the body varies with time we need to define its average speed.
Average speed is the total distance travelled by a body, divided by the total time taken to cover that distance Thus,
Average speed=\(\frac { Total\quad distance\quad travelled }{ Total\quad time\quad taken } \)
For example, if a car travels a distance of 10 km in 2 hours, then its
Average speed=\(\frac { 100\quad km }{ 2\quad hour } \) = 50 km per hour.
8.
| Distance | Displacement |
| 1.Distance is the length of the actual path traversed by a body irrespective of its direction of motion | 1. Displacement is the shortest distance between the initial and final position of a body in a given direction. |
| 2.Distance between two given points may be same or different for different paths chosen. | 2.Displacement between two given points is always same |
| It is a scalar quantity | It is a vector quantity |
| Distance covered is always positive or zero | Displacement covered may be positive negative or zero. |
9.
| Distance | Displacement |
| 1. Distance is the length of the actual path traversed by a body irrespective of its direction of motion | 1. Displacement is the shortest distance between the initial and final position of a body in a given direction. |
| 2. Distance between two given points may be same or different for different paths chosen. | 2. Displacement between two given points is always same |
| It is a scalar quantity | It is a vector quantity |
| Distance covered is always positive or zero | Displacement covered may be positive negative or zero. |
10.
Different types of Motions:
1.Translatory Motion When a body moves as a whole along a straight or curved path, it is said to be in translatory motion.It is again of two types:
(i) Rectilinear motion Here a body moves as a whole along a straight path.For example a train moving on straight rails has translatory rectilinear motion.
(ii) Curvilinear motion.Here a body moves as a whole along a curved path For example motion of a bicycle taking a turn along a curved path.
2.Rotatory motion.When a body rotates about a fixed point or axis , it has rotatory motion For example motion of a flywheel about a shaft.
3.Vibratory or oscillatory motion. When a body moves to and from about a mean position again and again it has vibratory or oscillatory motion For example the motion of the pendulum of a wall clock.
4.Complex motion. Sometimes ,the motion of a body may be a combination of more than one types of motion For example a ball rolling down an inclined plane has botth translatoruy and rotatory motions.
11.
a = 8 \(ms^{ -2 }\),v = 32 \(ms^{ -1 }\)
(ii) t = \(2\sqrt { 2\quad s } \)
12.
\(v_{ av }=\frac { s_{ 1 }+s_{ 2 } }{ t_{ 1 }+t_{ 2 } } =\frac { s }{ \frac { s_{ 1 } }{ v_{ 2 } } +\frac { s_{ 2 } }{ v_{ 2 } } } \)
\(40=\frac { 100 }{ \frac { 30 }{ 30 } +\frac { 70 }{ v_{ 2 } } } \)
\(=1+\frac { 70 }{ v_{ 2 } } =\frac { 100 }{ 40 } =\frac { 5 }{ 2 } \)
\(\frac { 7 }{ v_{ 2 } } =\frac { 5 }{ 2 } -1=\frac { 3 }{ 2 } \)
\(v_{ 2 }=\frac { 7X2 }{ 3 } =\frac { 14 }{ 3 } =4.67\quad km\quad h^{ -1 }\)
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