CENTRAL MODERN SCHOOL

LESSON PLAN FOR PHYSICS OF CLASS IX (Session : 2020 – 2021)

Continued from lessons sent in the previous week (i.e. , from 22-06-2020 to 26-06-2020)

For the week starting from 29-06-2020 to 03-07-2020

TOPIC : Motion In One Dimension

DAY 1

Solved Numericals on Graphical Representation

1. The figure below shows the displacement-time graph for the motion of two boys A and B along a straight road in the same direction.

Answer the following :

(i) When did B start after A?

(ii) How far away was A from B when B started?

(iii) Which of the two has greater velocity? Why?

(iv) When and where did B overtake A?

Solution :

(i) B started his motion 2 h later from the start of A.

(ii) When B started, A was at a distance of 10 km away from B.

(iii) B has greater velocity than A since the straight line on graph for B has greater slope than that for A.

(iv) B overtook A after travelling 20 km and after 2 h.

2. A car travels with a uniform velocity of 20 ms-1 for 5 s. The brakes are then applied and the car is uniformly retarded. It comes to rest in further 8 s. Draw a graph of velocity against time. Use this graph to find :

(i) the distance travelled in first 5 s,

(ii) the distance travelled after the brakes are applied,

(iii) the total distance travelled, and

(iv) acceleration during the first 5 s and last 8 s.

Solution :

The graph of velocity against time is shown below.

(i) The distance travelled in first 5 s = area of rectangle OABD

= OD × OA = 5 s × 20 ms-1

= 100 m

(ii) The distance by car after the brakes are applied = area of ∆ BDC

= × DC × DB

= × (13 – 5) × 20

= 80 m

(iii) Total distance travelled = area of rectangle OABD + area of triangle BDC

= 100 + 80 = 180 m

(iv) Acceleration in the first 5 s (in part AB) = 0

(since straight line AB is parallel to the time axis, so slope = 0).

Acceleration in the last 8 s (in part BC)

= Slope of line BC

= = ( )( )

=

= − 2.5 ms-2

Since acceleration is negative, so retardation = 2.5 ms-2.

4. A train starts from rest and accelerates uniformly at 100 m minute-2 for 10 minutes. It then maintains a constant velocity for 20 minutes. The brakes are

then applied and the train is uniformly retarded. It comes to rest in 5 minutes. Draw a velocity time-graph and use it to find :

(i) the retardation in the last 5 minutes,

(ii) total distance travelled, and

(iii) the average velocity of the train.

Solution :

The velocity-time graph is drawn below.

(i) Retardation in the last 5 minutes = − Slope of the line BC

= − = − ( )( )

= − = 200 m minute-2

(ii) Total distance travelled

= Area of trapezium OABC

= (OC + AB) × AD

= (35 + 20) × 1000

= 55 × 500

= 27500 m (or 27.5 km)

(iii) Average velocity =

= = 785.7 m minute-1

4. The figure below shows the velocity-time graphs for two cars A and B moving in same direction. Which car has greater acceleration? Give reason to your answer.

Solution :

Car B has greater acceleration.

Reason : Slope of line B is more than that of line A.

(Submission of exercise books for work done from 16th June, 2020 will be from 6th July, 2020 to 8th July, 2020 from 11 am to 1 pm)

(To be continued..............)

DAY 2

Equations of Motion

For motion of a body moving with a uniform acceleration, the following three equations give the relationship between initial velocity (u), final velocity (v), acceleration (a), time of journey (t) and distance travelled (S) :

(1) 푣 = 푢 + 푎푡

(2) 푆 = (푢 + 푣) 푡 = 푢푡 + 푎푡 and

(3) 푣2 = 푢2 + 2푎푆

For a falling body under gravity (‘g’ is positive)

(4) 푣 = 푢 + 푔푡

(5) 푆 = 푢푡 + 푔푡

(6) 푣2 = 푢2 + 2푔푆

For a body thrown up against gravity (‘g’ is negative)

(7) 푣 = 푢 − 푔푡

(8) 푆 = 푢푡 − 푔푡

(9) 푣2 = 푢2 − 2푔푆

Now, we have to derive the first three formula of motion

1. Derivation of 풗 = 풖 + 풂풕

By definition,

Acceleration =

=

or 푎 =

or 푎푡 = 푣 − 푢

or 푣 = 푢 + 푎푡

2. Derivation of 푺 = 풖풕 + ퟏퟐ 풂풕ퟐ

Distance travelled = Average velocity × time

= ( ) × time

or 푆 = × 푡

But, from, 푣 = 푢 + 푎푡

∴ 푆 = ( ) × 푡

or 푆 = (2푢 + 푎푡2 ) × 푡

or 푆 = 푢푡 + 푎푡

3. Derivation of 풗2 = 풖2 + ퟐ풂푺

Distance travelled = Average velocity × time

or 푆 = × 푡

But, from, 푣 = 푢 + 푎푡

or 푡 =

∴ 푆 = × =

or 푣 − 푢 = 2푎푆

or 푣2 = 푢2 + 2푎푆

Note : When a body starts from rest initial velocity (푢) = 0

When a body comes to rest final velocity (푣) = 0

When a body is thrown vertically upwards velocity at the highest point is zero (푣 = 0).

When a body falls from a certain height then initial velocity is zero (푢 = 0).

(Submission of exercise books for work done from 16th June, 2020 will be from 6th July, 2020 to 8th July, 2020 from 11 am to 1 pm)

(To be continued..............)

DAY 3

Solved Numericals on Motion

1. A body, initially at rest, starts moving with a constant acceleration 2 ms-2. Calculate : (i) the velocity acquired and (ii) the distance travelled in 5s.

Solution : 푢 = 0 , 푎 = 2 ms-2 , 푡 = 5 푠 , 푣 = ? , 푆 = ?

(i) 푣 = 푢 + 푎푡

⇒ 푣 = 0 + 2 × 5

⇒ 푣 = 10

∴ velocity acquired is 10 ms-1

(ii) 푆 = 푢푡 + 푎푡

⇒ 푆 = 0 + × 2 × 5 × 5

⇒ 푆 = 25

∴ distance travelled is 25 m.

2. A body moves from rest with a uniform acceleration and travels 270 m in 3 s. Find the velocity of the body at 10 s after the start.

Solution : 푢 = 0 , 푆 = 270 m , 푡 = 3 s , 푎 = ?

푆 = 푢푡 + 푎푡

⇒ 270 = 0 + × 푎 × 3 × 3

⇒ 푎 = ×

⇒ 푎 = 60

∴ acceleration is 60 ms-2

Now, velocity of a body after 10 s is

푣 = 푢 + 푎푡

⇒ 푣 = 0 + 60 × 10

⇒ 푣 = 600

∴ final velocity is 600 ms-1

3. A train is moving with a velocity of 90 km h-1. It is brought to stop by applying the brakes which produce a retardation of 0.5 ms-2. Find : (i) the velocity after 10 s, and (ii) the time taken by the train to come to rest.

Solution : 푢 = 90 km h-1 = 90 × ms-1 = 25 ms-1

푣 = 0 , 푎 = −0.5 ms-2 , 푣 = ? , 푡 = ?

(i) velocity after 10 s is

푣 = 푢 + 푎푡

⇒ 푣 = 25 + (0.5 × 10)

⇒ 푣 = 25 − 5 = 20

∴ velocity of the train after 10 s is 20 ms-1.

(ii) 푣 = 푢 + 푎푡

∴ 0 = 25 + (−0.5 × 푡)

⇒ −0.5 × 푡 = 25

⇒ 푡 = − .

= −

⇒ 푡 = −50

Since time cannot be negative, so 푡 = 50 s.

Now, answer the following questions in fair copy.

Q1. Derive the formula : 푣 = 푢 + 푎푡 Q2.

The figure above shows the displacement of a body at different times. From the figure :

(a) Calculate the velocity of the body as it moves for time interval (i) 0 s to 5 s, (ii) 5 s to 7 s and (iii) 7 s to 9 s.

(b) Calculate the average velocity during the time interval 5 s to 9 s.

Q3. The velocity-time graph of a moving body is given below.

From the figure find :

(i) the acceleration in parts AB, BC, CD.

(ii) displacement in each part AB, BC, CD.

(iii) total displacement.

Q4. Draw a displacement-time graph for a boy going to school with uniform velocity.

(Submission of exercise books for work done from 16th June, 2020 will be from 6th July, 2020 to 8th July, 2020 from 11 am to 1 pm)

(Explanation of the topic completed)