You can Download Equations of Motion Questions and Answers, Summary, Activity, Notes, Kerala Syllabus 9th Standard Physics Solutions Part 1 Chapter 2 help you to revise complete Syllabus and score more marks in your examinations.

## Kerala State Syllabus 9th Standard Physics Solutions Chapter 2 Equations of Motion

### Equations of Motion Textual Questions and Answers

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.

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:

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

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.

Question 5.

Using the graph find out the displacement of the car in 3 s.

Answer:

Displacement = 11 m

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 2^{nd} and 8^{th} 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

= t^{2} – t^{1}

AQ = QR – AR

= v- u

Formatin of equation showing velocity – time relation.

Acceleration = \(\frac{\text { Change in velocity }}{\text { time }}\)

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 + \(\frac { 1 }{ 2 }\) at^{2} 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.

v^{2} = u^{2} + 2as

Equations of motion

v = u+at

s = ut + \(\frac { 1 }{ 2 }\) at^{2}

v^{2} = u^{2} + 2as

Question 7.

The velocity of a body starting from rest is 20 m/s in the 4^{th} second and 40 m/s in the 8^{th} second. What is the distance traveled by the body between the 4^{th} and 8^{th} second?

Answer:

Velocity at the 4^{th} second u = 20 m/s

Velocity at the 8^{th} second v = 40 m/s

Question 8.

A car came to rest when brake was applied for 4s to get a retardation of 3 m/s^{2}. Calculate how far the car would have traveled after applying the brake.

Answer:

a = -3 m/s^{2}

t = 4s

v = 0

v = u+at

-u = -3 × 4 + 0

u =12 m/s

Displacement of the car

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

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

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=?

Question 12.

An object starting from rest travels with an acceleration of 5 m/s^{2}. What will be its velocity after 3 s?

Answer:

u=0,

a= 5m/s^{2},

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/s^{2})

A → B Uniform retardation (1m/s^{2})

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/s^{2}

s = ut + 1/2 at^{2}

= 5 × 5 + 1/2 × 2 × 5^{2}

= 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 at^{2}

v^{2} = u^{2} + 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/s^{2}. Calculate the velocity and distance traveled after 1 minute?

Answer:

v = u + at

u = 0

= 0+ 5m/s^{2} x 60 s

a = 5m/s^{2}

= 0+ 300 m/s

t = 1

mt = 60 sec

= 300 m/s

s =ut + 1/2 at^{2}

= 0 × 60 + 1/2 × 5 × 60^{2}

= 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