BABITA SHARMA (ROLL NO.5)

                                        ch-3 :- MOTION 

1) Define the scalar quantity and name 5 physical quantities which are scalar and give reasons why these physical quantities are scalar?

ANS- Scalar quantity is the physical quantity in which only magnitude is measured and                       direction is not specified .

 

 

 

Graph of the above table:-

                                                            

It is a non-uniform motion.

 

 

    EXERCISE

 
1. 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?

Number of round in 40 s =1 round
Number of round in 140 s =140/40
=3 ¹/₂

Answer

Diameter of circular track (D) = 200 m
Radius of circular track (r) = 200 / 2=100 m
Time taken by the athlete for one round (t) = 40 s
Distance covered by athlete in one round (s) = 2π r
= 2 x ( 22 / 7 ) x 100
Speed of the athlete (v) = Distance / Time
= (2 x 2200) / (7 x 40)
= 4400 / 7 × 40
Therefore, Distance covered in 140 s = Speed (s) × Time(t)
= 4400 / (7 x 40) x (2 x 60 + 20)
= 4400 / ( 7 x 40) x 140
= 4400 x 140 /7 x 40
= 2200 m

2. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s⁻² for 8.0 s. How far does the boat travel during this time?

Answer

Given Initial velocity of motorboat,  u = 0
Acceleration of motorboat,  a = 3.0 m s^-2
Time under consideration,  t = 8.0 s
We know that Distance, s = ut + (1/2)at²
Therefore, The distance travel by motorboat = 0 x 8 + (1/2)3.0 x 8 2
= (1/2) x 3 x 8 x 8 m
= 96 m 

3.An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Answer

Radius of the circular orbit, r= 42250 km
Time taken to revolve around the earth, t= 24 h
Speed of a circular moving object, v= (2π r)/t
=[2× (22/7)×42250 × 1000] / (24 × 60 × 60)
=(2×22×42250×1000) / (7 ×24 × 60 × 60) m s^-1
=3073.74 m s ^-1

4. State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity.
(b) an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer

(a)
When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s2.

(b)
When a car is moving in a circular track, its acceleration is perpendicular to its direction.

7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s⁻², with what velocity will it strike the ground? After what time will it strike the ground?

Answer

Let us assume, the final velocity with which ball will strike the ground be 'v' and time it takes to strike the ground be 't'
Initial Velocity of ball,  u =0
Distance or height of fall,  s =20 m
Downward acceleration,  a =10 m s^-2
As we know, 2as =v²-u²
v² = 2as+ u²
= 2 x 10 x 20 + 0
= 400 
∴ Final velocity of ball, v = 20 ms^-1
t = (v-u)/a
∴Time taken by the ball to strike = (20-0)/10
= 20/10
= 2 seconds