Kerala Syllabus Class 10 Biology Solutions Chapter 3 Chemical Messages for Homeostasis

 Kerala Syllabus 9th Standard Physics Notes Chapter 2 Question 1.

Draw Position – Time graph. What is the nature of the grap?

X-axis, time (s)	0	1	2	3
Y-axis, position (m)	0	1	2	3

Answer:

The graph is straight line.

Class 9 Physics Chapter 2 Kerala Syllabus Question 2.
Using the data given below, draw a position-time graph

Time (s)	0	1	2	3	4	5	6
Position (m)	0	2	4	6	8	10	12

Answer:

Physics Class 9 Chapter 2 Equations Of Motion Kerala Syllabus Question 3.
The position-time graph regarding the motion of a car is given. Find out from the graph the distance traveled by the car in 8 s.

Answer:
Draw a perpendicular to the position-time graph at the eighth second. From the point where the perpendicular meets the graph, draw another perpendicular to the Y-axis. The point at which this perpendicular meets the Y-axis is the distance traveled by the car in 8 sec. ie 40m.
The distance covered by the car is 8s = 40 m

9th Class Physics Chapter 2 Equations Of Motion Kerala Syllabus Question 4.
Complete the table using the data from the following position-time graph related to the motion of a car.

Position (m)	0	2	4	6	8	10
Velocity (m/s)	0	10	20	30	40	50

Change the scale and draw another graph.

Displacement of the body in 5s from both graphs is 25 m
We use scales while drawing a graph, to contain the given measurements on a graph paper. The size of the graph decreases as we increase the scale considered. But the value doesn’t change.
Motion of a car is shown below using a diagram.

Complete the table with the help of the figure

Position (m)	0	2	4	6	8	10
Velocity (m/s)	0	5	20	36	44	50

Position – time graph of a body moving with uniform speed will be a stright line. The body will be moving with nonuniform speed when the graph is not in a straight line.

9th Class Physics Notes Kerala Syllabus  Question 5.
Using the graph find out the displacement of the car in 3 s.
Answer:
Displacement = 11 m

Kerala Syllabus 9th Standard Physics Notes Question 6.3
From the graph find out the time taken by the car to travel a distance of 45 m.
Answer:
Time = 8.4s

Velocity – Time Graph

Time (s)	0	2	4	6	8	10
Velocity (m/s)	10	10	10	10	10	10

With the given data in the table, draw a velocity-time graph.

The displacement of the object between the 2nd and 8th second
= Velocity × time
= AB × AD = 10 × 6 = 60 m
= Area of ABCD
= Area of the portion under the part BC of the graph
Displacement of a body within a definite interval of time is equal to the area of the portion under velocity-time graph.

Equations Of Motion

The velocity-time graph of an object traveling with uniform acceleration (freely falling stone) is given below.

PS = AR
= u
QR = v
= t2 – t1
AQ = QR – AR
= v- u
Formatin of equation showing velocity – time relation.
Acceleration =  Change in velocity / time 

v- u = at
v = u + at.
This is the first equation of motion.
Formation of equation showing Position- Time Relation
Displacement = area of trapezium PQRS

S = ut + 1/2 at2 this is known as second equation of motion.
Formation of equation showing Position – Velocity Relation
Displacement = The area of the quadrilateral

This equation helps us to calculate the final velocity v of an object using u, a and s, even if the time taken to travel is unknown.
v2 = u2 + 2as
Equations of motion
v = u+at
s = ut + 1/2 at2
v2 = u2 + 2as

9th Class Physics Equations Of Motion Kerala Syllabus Question 7.
The velocity of a body starting from rest is 20 m/s in the 4th second and 40 m/s in the 8th second. What is the distance traveled by the body between the 4th and 8th second?
Answer:
Velocity at the 4th second u = 20 m/s
Velocity at the 8th second v = 40 m/s

9th Physics Notes Kerala Syllabus Question 8.
A car came to rest when brake was applied for 4s to get a retardation of 3 m/s2. Calculate how far the car would have traveled after applying the brake.
Answer:
a = -3 m/s2
t = 4s
v = 0
v = u+at
-u = -3 × 4 + 0
u =12 m/s
Displacement of the car

Kerala Syllabus 9th Standard Physics Notes Chapter 1 Question 9.
If the velocity of a car moving with uniform velocity changes from 20 m/s to 40 m/s in 5s
a) What is the acceleration of the car?
b) What is the displacement by the car during this time interval?
Answer:
a) u = 20 m/s,
v = 40 m/s,
t= 5 s

Kerala Syllabus 9th Standard Physics Notes English Medium Question 10.
If the velocity of a train starting from rest becomes 72 km/h in 10 minutes.
a) What is the acceleration?
b) Calculate the distance traveled by the train within this time interval
Answer:
a) Here , u = 0,
v= 72 km/h

Kerala Syllabus 9th Standard Physics Notes Chapter 2 Malayalam Medium Question 11.
A car starting from rest travels 100 m in 5s with uniform acceleration. Find the acceleration of the car.
Answer:
u = 0,
t= 5s,
s = 100 m,
a=?

Kerala Syllabus 9th Standard Physics Textbook Solutions Question 12.
An object starting from rest travels with an acceleration of 5 m/s2. What will be its velocity after 3 s?
Answer:
u=0,
a= 5m/s2,
t= 3s,
v=?
v= u + at,
v= 0 + 5 × 3,
v= 15 m/s
Complete the table using the graph shown below. Compare.


Answer:

In a velocity-time graph when we draw perpendiculars to the graph from the time specified, we get a geometrical shape. The area of this geometrical shape will give displacement in the specified time interval.

Let Us Assess

Question 1.
Draw position-time graph

Time (s)	0	3	6	9	12	15	18
Position (m)	0	5	10	15	20	25	30

Answer:

Question 2.
Draw speed – time graph

Time (s)	0	2	4	6	8	10
Speed (m/s)	10	15	20	20	20	15

Answer:

Question 3.
Examine the graph and answer the following questions.

a) Is the motion of the object uniform/nonuniform?
b) Say whether the acceleration from O to A is uniform? What about from A to B?
Answer:
a) Non- uniform
b) O → A. uniform acceleration (0.4 m/s2)
A → B Uniform retardation (1m/s2)

Question 4.
Complete the table by analyzing the graph


Answer:

Position of the object in the graph	Nature of the motion
From A to B	Velocity increase
From B to C	Uniform velocity
From C to D	Velocity decreases

Question 5.
If a velocity of a train which starts from rest is 72 km/h (20 m/s) after 5 minutes, find out its acceleration and the distance traveled by the train in this time.
Answer:

Question 6.
A car attains a velocity of 54 km/h (15 m/s) within 5 seconds from an initial velocity of 18 km/h (5m/s). Calculate its acceleration and displacement.
Answer:
u = 5 m/s .
t = 5s
v = 15 m/s

= 2 m/s2
s = ut + 1/2 at2
= 5 × 5 + 1/2 × 2 × 52
= 25 + 25
= 50 m

Question 7.
Analyze the graphs given below.

a) Which graph indicates uniform velocity?
b) Which graph indicates nonuniform acceleration?
c) Which graph indicates the motion of freely falling stone?
Answer:
a) graph (2)
b) graph (4)
c) graph (3)

Equations of Motion More Questions and Answers

Question 1.
Write the equations of motion. What does each letter indicate?
Answer:
v = u + at
s = ut +1/2 at2
v2 = u2 + 2as
Where
u – Initial velocity
v – final velocity
a – acceleration
t – time
s – Distance (Displacement)

Question 2.
An object starting form rest travels with a uniform acceleration of 5 m/s2. Calculate the velocity and distance traveled after 1 minute?
Answer:
v = u + at
u = 0
= 0+ 5m/s2 x 60 s
a = 5m/s2
= 0+ 300 m/s
t = 1
mt = 60 sec
= 300 m/s
s =ut + 1/2 at2
= 0 × 60 + 1/2 × 5 × 602
= 9000 m

Question 3.
Draw a velocity-time graph on the basis of the given table.

Time (s)	0	5	10	15	20	25	30	35	40
Velocity(m/s)	0	10	20	20	20	15	10	5	0

Answer:

X – axis – 1cm = 5s
Y – axis – 1cm = 5 m/s

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