Textual Questions of motion (9th class 1st lesson)

chapter 1:-Motion

motion textual questions

Ans:-
She moves with uniform velocity.

Ans:-
1. Speed = distance/time
        = slope of the graph distance, time graphs
2. Slope of distance time graphs represents speed of an object
3. Curve of car 'A' has more slope than 'B'. So, car ‘A’ moves fast.

Ans:-

Speed	Velocity
1. The distance travelled by the body in unit time is called speed.	1. The displacement of the body in unit time is called velocity.
2. Speed = Distance / Time.	2. Velocity = Displacement / Time.
3. It is a scalar.	3. It is a vector.
4. It's value is always positive or zero.	4. It's value may be positive or zero or negative.
5. It indicates how fast the position of the object is changing.	5. It indicates how fast the position of the object is changing in a particular direction.

Ans:-
1. Rate of change in velocity is called acceleration.
2. If a body travels in a straight line and its velocity changes (increase or decrease) by equal amount in equal intervals of time, means the rate of change in velocity is constant, Then the acceleration is said to be uniform acceleration or constant acceleration.

Ans:-
(i) if a body moves in a curved path, it changes it's direction continuously. physical quantity which Changes it’s direction is a scalar.
(ii) As velocity is a vector quantity, we should not use it.
(iii) Instead of it we should use the term speed. Because speed is a scalar.
(iv) The correct statement is “ The car rounded the curve at a constant speed of 70 km/h.

Ans:-
(i) As velocity is a vector quantity, Constant velocity means rate of change in displacement and it's direction are constant.
(ii) it covers equal amount of displacement in equal intervel of time in a particular direction, it is same for longer and shorter time periods
(iii) If velocity is constant, the average velocity over any time interval is equal to instantaneous velocity at any time.
     V _average =V _constant

Ans:-
(i) In the case of a vertically projected body, at the maximum height of the body the velocity is zero and the acceleration of the body works towards the earth with a constant value 9.8 m/sec².
(ii) later it starts to fall freely, which means its velocity was reversed due to change in direction.
But its acceleration is the same as before, Acceleration due to gravity is 9.8 m/s² (constant)

Ans:-
(i) Displacement is the shortest distance in a specified direction.
(ii) Displacement vector from A to B is denoted with AB ⃗

Ans:- The distance – time graph for a body which is moving with speed decreasing gradually is as follows.

Ans:- (i) Hare and tortoise start at “O”.
(ii) They reached their final destination at different times.

(iii) OC represents the motion of the tortoise.
(iv) OABD represents the motion of hare.

Ans:-
Part 1: distance = 100 m
Time = 4 s
Average speed = Total distance/ Time taken
= 100/4
= 25 m/s

Case(ii): distance = 50 m
Time = 2 s
Average speed =Total distance/ Time taken
= 50/2 = 25 m/s

Ans:-

Ans:-
A stone is dropped into the well.
Initial velocity U = 0 m/s
Acceleration due to gravity (g) = 10 m/s²
Time taken to reach the water surface = 2s
Velocity of the stone while it touches the water surface V = ?

Formula :- V = U + at
V = U + gt
V = 0 + 10(2)
V = 20 m/s
--------------------------------------------------
Depth of water surface from the top of well S = ?
Formula : S=Ut+1/2 at²
S = 0(2) +1/2(10)(2)²
S = (10)(2)
S = 20m

Ans:-
Given
Initial velocity U = 6 m/s
Acceleration = 2 m/s²
Timet = 3s
Distance travelled by the object S = ?
Formula :- S=Ut+1/2 at²
S = 6(3) +½ (2)(3)²
S = 18 + 9
S = 27m

Ans:-
Given
Distance travelled by the car S = 8 m
Initial velocity U = 40 m/s
Final velocity V = 0 m/s (since Car stopped due to applying brakes.)
Accelerationa = ?
Formula : v2 - u2 = 2as
02 – (40)2 = 2a(8)
- 1600 = 16 a
- 100 = a
a = - 100m/s²
-ve sign indicates retardation ( deceleration )
So, deceleration = 100m/s²

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