JKBOSE 9th Class Science Solutions Chapter 1 Motion
JKBOSE 9th Class Science Solutions Chapter 1 Motion
JKBOSE 9th Class Science Solutions Chapter 1 Motion
Jammu & Kashmir State Board JKBOSE 9th Class Science Solutions
J&K class 9th Science Motion Textbook Questions and Answers
BASIS AND BASICS
◆ A body is said to be in motion when it changes its position with time.
◆ At any instant a body may appear to be in motion to one person but the same object may appear at rest to another person.
◆ To describe the position of an object we are required to fix directive point which is called origin.
◆ The motion of some objects can be controlled while that of some other objects remains uncontrolled and irregular.
◆ When an object moves along a straight line path then its motion is called linear motion.
◆ Those quantities which can be described completely by their magnitude only are called physical quantities.
◆ The shortest distance measured between initial and final position is called displacement of the body.
◆ When the final position of the body in motion coincides with its initial position then the value of displacement becomes equal to zero.
◆ The displacement of an object may be zero but its distance will not be zero.
◆ Odometer is a device which measures the distance covered by automobiles.
◆ If an object covers equal distances in equal intervals of time however small the
interval may be then its motion is called uniform motion.
◆ If a moving object covers unequal distances in equal intervals of time then the motion of the object is called non-uniform motion.
◆ The rate of change of motion of an object is called its speed. The unit of speed is ms-¹
◆ The average speed of an object is obtained by dividing distance covered by the body by time taken to cover that distance.
∴ Average speed = Total distance covered / Time taken to cover that distance
◆ The rate of change of motion of a body in a particular direction is called its velocity i.e. speed in a particular direction is called velocity.
◆ If the velocity of an object changes by same rate then its mean (average) velocity, is obtained by taking arithmatic mean of initial velocity and final velocity.
Average Velocity = Initial velocity + Final velocity / 2
◆ Time interval is measured by digital wrist watch or stop watch.
◆ Speed of sound in air = 340 m/s.
◆ Speed of light in air = 3 x 10⁸ m/s.
◆ The rate of change of velocity of a body is called its acceleration
Acceleration = Change in velocity / Time taken
S.I unit of acceleration is m s–².
IMPORTANT TERMS/FACTS TO MEMORISE
⇒ Rest. A body is said to be at rest if it does not change its position with respect to its surroundings.
⇒ Motion. A body is said to be in motion if it changes its position with time with respect to the surroundings.
⇒ Distance. It is the length of the actual path travelled by a body between its initial position and final position.
⇒ Displacement. The change in position of an object in a given direction is known as displacement. It is measured by the shortest distance moved by a body from the initial position to final position.
⇒ Uniform Motion. If an object covers equal distances in equal intervals of time however small the time intervals may be, then the motion of the object is said to be uniform motion.
⇒ Non-Uniform Motion. If an object covers unequal distances in equal intervals of time then the object is said to be in non-uniform motion.
⇒ Speed. The distance travelled by a body in a unit time.
Speed = Distance travelled / Total time taken
⇒ Uniform Speed. When an object covers equal distances in equal intervals of time, however small these intervals may be, it is said to move with uniform speed.
⇒ Non-Uniform Speed. When an object covers unequal distances in equal intervals of time, it is said to move with non-uniform speed.
⇒ Average Speed. It is the total distance travelled by an object divided by the time taken to travel that distance.
Average speed = Total distance travelled / Total time taken
⇒ Velocity. It is the displacement produced per unit time.
Or
It is the distance travelled per unit time in a given direction.
Velocity = Displacement / Time
⇒ Uniform Velocity. If a body covers equal distances in equal intervals of time, however small these intervals may be, in a particular direction, its velocity is said to be uniform velocity.
⇒ Variable Velocity. If a body travels unequal distances in equal intervals of time of changes its direction or both then its velocity is said to be variable velocity.
⇒ Average Velocity. It is defined as the net displacement covered divided by the total time taken.
⇒ Scalar Quantities. Those physical quantities which require only magnitude for their complete description, are called scalar quantities.
⇒ Vector Quantities. Those physical quantities which need both magnitude and direction for their complete description are called vector quantities.
⇒ Acceleration. It is defined as the rate of change of velocity with time.
Acceleration = Change of velocity / Total Time Taken
⇒ Uniform Acceleration. A body is said to travel with uniform acceleration if its velocity changes by equal amounts in equal intervals of time, however small these intervals may be.
TEXTBOOK QUESTIONS (SOLVED)
Q. 1. An object has moved through a distance. Can it have zero displacement ? If yes, support your answer with an example.
Ans.— Yes, a body can have zero displacement, if this body while moving occupies its final position coinciding with its initial position.
Example. Suppose a body starting its motion from initial position O covers some distance and reaches a position A. If this body while moving returns to its initial position O than in that situation its displacement will be zero.
But distance covered by the body = OA + AO
= 60 km + 60 km
= 120 km
Q. 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds ?
Sol.— Total distance round the boundary of field once (i.e. circumference)
= AB+BC + CD + DA
= 10m + 10m + 10m + 10m
= 40m
Time taken to go round the field once = 40 s
Total time taken = 2minutes 20seconds
= (2 x 60 + 20) seconds
= (120 + 20) seconds
= 140 seconds
Time taken by farmer to complete 3 rounds of field
= 3 x 40 s
= 120 s
Time left after completing 3 rounds of field = (140 – 120)s
= 20 s
∴ Distance covered by farmer in 40 s = 40m
∴ Distance covered in 1 s = 1m
Distance that would be covered in 20 s = 20m
In other words farmer starting from point A and while going along the boundary of the field and after completing 3 rounds in 2 min 20 s would reach the point C.
∴ Displacement = AC
(the shortest distance between initial and final position)
= √AB² + BC²
= √(10)² + (10)²
= √100 + 100
= √ 200 m
= √ 100 x 2 m
= 10√2 m
= 10 x 1.414 m
= 14.14 m Ans.
Q. 3. Which of the following is true for displacement ?
(a) It cannot be zero
(b) Its magnitude is greater than distance travelled by the object.
(c) Its magnitude is less than or equal to distance travelled by the object.
Ans.— (c) Its magnitude is less than or equal to distance travelled by the object.
Q. 4. Distinguish between speed and velocity.
Or
Give two points of difference between speed and velocity.
Ans.— Distinction between speed and velocity :
| S.NO | Speed | Velocity |
| 1. | It is defined as the rate of a change of a position of a Body i.e. the distance covered by a body per unit time. |
It is defined as the rate of change of displacement of a Body. i.e. it is the speed in a particular direction.
|
| 2. | It is a Scalar quantity and can be completely represented by its magnitude only. | It is a vector quantity. To represent it completely it requires both magnitude and direction. |
| 3. | Speed of an object is always positive. | Velocity of an object can be both positive and negative. |
Q. 5. Under what condition(s) is the magnitude of average velocity of an object is equal to its average speed ?
Ans.— We know, Average speed = Total distance travelled / Total time taken
and Average velocity = Displacement / Total time
When a body travels in a straight line with variable motion in the same direction then total distance covered and displacement are equal in magnitude. In this case the average speed and average velocity are equal.
Q. 6. What does the odometer of an automobile measure ?
Ans.— The odometer of an automobile measures the distance covered by it.
Q. 7. What does the path of an object look like when it is in uniform motion ?
Ans.— When an object is in uniform motion, it moves along a straight line. But an object can also move with uniform motion along a circular path.
Q. 8. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at a speed of light that is 3 x 10⁸ m s-¹.
Sol.— Time taken by the signal to reach the ground station from spaceship
(t) = 5 min
= 5 x 60 s
= 300
Speed of signal (v) = Speed of light = 3 x 10⁸ ms–¹
Distance of the spaceship from earth (s) = ?
Distance of spaceship from ground (s)
= speed of signal (v) x Time (t)
= 3 x 10⁸ x 300
= 3 x 10⁸ x 3 x 10²
= 9 x 10⁸ x 10²
= 9 x 10¹⁰ m Ans.
Q. 9. When will you say a body is in:
(i) Uniform acceleration ?
(ii) Non-uniform acceleration ?
Ans.—(i) Uniform acceleration. When a body travels in a straight line and its velocity changes by equal amounts in equal intervals of time then it is said to travel with uniform acceleration.
(ii) Non-uniform acceleration. When the velocity of a body changes by unequal amounts in equal intervals of time then the body is said to travel with non-uniform acceleration.
Q. 10. A bus decreases its speed from 80 km h-¹ to 60 km h-¹ in 5 s. Find the acceleration of the bus.
Sol.— Initial velocity of bus (u) = 80 km h-¹
= 80 x 5/18 m s-¹
Final velocity of bus (v) = 60 km h-¹
= 60 x 5/18 m s-¹ [∴ 1 km h -¹ = 5/18m s-¹]
Time taken (t) = 5 s
Acceleration of the bus (a) = ?
We know, acceleration (a) Change in velocity / Time taken
= Final velocity (v) — Initial velocity (u) / Time taken (t)
= 60 x 5 / 18 – 80 x 5 / 18 / 5
= 5 / 18 (60 – 80) / 5
= 5 / 18 (-20) / 5
= -100 / 18 x 1 / 5
= -10 / 9 m s-²
= -1.11 m s-² Ans.
Hence, the bus has negative acceleration (retardation).
Q. 11. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h-¹ in 10 minutes. Find its acceleration.
Sol.— Initial velocity of train (u) = 0 [because the train starts from rest)
Final velocity of train (v) = 40 km h-¹
= 40 x 5/18 ms-¹
= 100 / 9 ms-¹
Time taken (t) = 10 minutes
= 10 x 60 s
= 600 s
Acceleration (a) = ?
We know, acceleration (a) = Change in velocity / Time taken
= Final velocity (v) — initial velocity (u) / Time taken (t)
= v – u / t
= 100 / 9 – 0 / 600
= 100 / 9 x 1 / 600
= 1 / 54 ms-²
= 0.018 ms-² Ans.
Q. 12. What is the nature of the distance – time graphs (x – t) for uniform and non – uniform motion of an object ?
Ans.— When a body covers equal distances in equal intervals of time, then it is said to travel with uniform motion. In w.w this situation, the distance covered by the body is directly proportional to the time taken. Therefore, distance-time (x-t) graph for uniform motion is a straight line.
Distance – time (x – t) graph for nonuniform motion may be a curved graph of any shape because a body travels unequal distances in equal intervals of time.
Q. 13. What can you say about the motion of object whose distance – time graph is a straight line parallel to time axis ?
Ans.— The object whose distance – time (x – t) graph is a straight line parallel to the time axis will be at rest with respect to the surrounding.
Q. 14. What can you say about the motion of an object if it’s speed – time graph is a straight line parallel to time axis ?
Ans.— The object whose speed – time (v – t) graph is a straight line parallel to time axis shows that it is in motion with uniform speed.
Q. 15. What is the quantity which is measured by the area occupied below velocity – time graph ?
Ans.— The area occupied below velocity – time graph measure distance covered by the body.
Q.16. A bus starting from rest moves with a uniform acceleration of 0.1 ms-² for two minutes. Find (a) the speed acquired (b) the distance travelled.
Sol.— (a) Initial speed of the bus (u) = 0 (Starting from Rest)
Acceleration of the bus (a) = 0.1ms-²
Time taken (t) = 2 minutes
= 2 x 60 s
= 120 s
Final speed of the bus (v) = ?
Distance travelled by the bus (S) = ?
We know, v = u + at
v = 0 + 0.1 x 120
v = 1 x 12
v = 12 ms-¹ Ans.
(b) Again, using S = ut + 1/2 at²
S = 0 x 120 + 1/2 x 0.1 x (120)²
= 0 + 1/2 x 0.1 x 120 x 120
= 720 m Ans.
Q. 17. A train is travelling at a speed of 90 km h-¹. Brakes are applied so as to produce a uniform acceleration of -0.5 ms-². Find how far the train will move before it is brought to rest ?
Sol.— Initial speed of train (v) = 90 km h-¹
= 90 x 5/18 ms-¹
= 5 x 5 ms-¹
= 25 ms-¹
Uniform acceleration (a) = -0.5 ms-²
Final speed of the train (v) = 0 (∴ the train is brought to rest)
Distance moved by the train (S) = ?
We know, v² – u² = 2aS
(0)² – (25)² = 2 x (-0.5) x S
-25 x 25 = -1 x 5
∴ S = 625 m Ans.
Q. 18. A trolley, while going down an inclined plane has an acceleration of 2 cms-². What will be its velocity 3 s after the start ?
Sol.— Here intial velocity of trolley (u) = 0 [∴ starting from rest]
Acceleration (a) = 2 cm s-²
Time (t) = 3 s
Final velocity of trolley (v) = ?
We know, v = u + at
v = 0 + 2 x 3
∴ Final velocity of trolley (v) = 6 cm s-¹ Ans.
Q. 19. A racing car has uniform acceleration of 4 ms-². What distance will it cover in 10 s after start ?
Sol.— Acceleration of racing car (a) = 4 m s-¹
Initial velocity of racing car (u) = 0
Time (t) = 10 s
Distance covered by the car (S) = ?
We know, S = ut + 1/2 at²
S = 0 x 10 + 1/2 x 4 x (10)²
S = 0 + 2 x 10 x 10
∴ Distance covered by racing car (S) = 200 m Ans.
Q. 20. A stone is thrown in a vertically upward direction with a velocity of 5 m s-¹. If the acceleration of the stone during its motion is 10 m s-² in the downward direction. What will be the height attained by the stone and how much time will it take to reach there ?
Sol.— Here, initial velocity (u) = 5 m s¹
Acceleration (a) = -10 m s-²
[∴ it moves upward against the gravity]
Final velocity of stone (v) = 0 [At the highest point it is brought to rest]
Height attained (S = h) = ?
Time taken (t) = ?
We know, v = u + at
0 = 5 + (-10) x t
0 = 5 – 10 x t
Or, t = 5/10
∴ Time taken (t) = 0.5 s Ans.
Again, using, v² – u² = 2aS
(0)² – (5)² = 2 x -10 x h
-5 x 5 = – 20 x h
Or, h = -25 / -20
= 5/4
∴ Height attained (h) = 1.25 m Ans.
TEXTBOOK ADDITIONAL EXPENSES (SOLVED)
Q. 1. An athlete completes one round of circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 min 20 s ?
Sol.— Diameter of circular track (d) = 200 m
Radius of circular track (r) = d / 2 = 200m / 2
= 100 m
Length of circular track (circumference) = 2 πr
= 2 x 22/7 x 100
= 4400 / 7 m
Time taken to complete 1 revolution (t) = 40 s
Total time = 2 minutes 20 seconds
= (2 x 60 + 20) seconds
= 140 s.
Distance covered in 40 s = 4400 / 7 m = (Circumference of 1 complete circular track)
Distance covered in 1 s = 4400 / 7 x 40 m
Distance covered in 140 s = 4400 / 7 x 40 x 140
= 2200 m Ans.
= 1/2 length of circular track.
An athlete starting from A and going in clockwise direction returns to point A in 3 rounds and reaches point B in 3.5 rounds.
∴ Displacement in 3.5 rounds= AB = shortest distance between initial and final position = 200 m from A to B.
Q. 2. Joseph jogs from end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B (b) from A to C ?
Sol.— (a) Length between end point A and end point B (AB) = 300 m
Time taken (t) = 2 min. 30 s
= (2 x 60 + 30) s
= (120 + 30) s
= 150 s.
Average speed = Average velocity
= Total distance between A and B (AB) / Total time (t)
= 300 m / 150 s
= 2 ms-¹ Ans.
(b) Length from end A to end B + Length on return from B to point C.
= AB + BC
= 300 m + 100 m
= 400 m
Total Time = 2 min 3p s + 1 min
= 3 min 30 s
= (3 x 60 + 30) s
= (180 + 130) s
= 210 s
Average speed = 400 m / 210 s
= 40 / 21 ms-¹
= 1.9 ms-¹ Ans.
Average Velocity = (300 – 100) m / 210 s
= 200m / 210s
= 20/21 ms-¹
= 0.95 ms-¹ Ans.
Q. 3. Abdul while driving to school, computes the average speed for his trip to be 20 km h-¹. On his trip along the same route, there is less traffic and average speed is 40 km h-¹. What is the average speed for Abdul’s trip ?
Sol.— Suppose distance of school from home = L
Distance of school from home and from school to home = L + L = 2L
Average speed while going to school (v₁) = 20 kmh-¹
Average time taken while going to school (t1) = Distance/Average speed
= L / v1
Average speed while returning from school to home (v2) = 30 kmh-¹
∴ Time taken in returning from school to home (t2) = L / v2
Total time taken complete journey = t1 + t2
= L/v1 + L/v2
= v2 x L + v1 x L / v1 x v2
= L(v2 + v1) / v1 x v2
Average speed = 2L / L (v1 + v2/ v1 x v2)
= 2L x v1 x v2 / L (v1 + v2)
= 2v1 v2 / v1 + v2
= 2 x 20 x 30 / (20 + 30)
= 2 x 20 x 30 / 50
= 24 km h-¹ Ans.
Q. 4. A motorboat starting from rest on a like accelerates in a straight line at a constant rate of 3.0 ms-² for 8.0 s. How far does the boat travel during this time ?
Sol.— Here, initial velocity of motorboat (u) = 0 [Starting from rest] Acceleration (a) = 3.0 ms-²
Time (t) = 8.0 s
Distance covered by the motorboat (s) = ?
We know, S = ut + 1/2 at²
= 0 x 8 + 1/2 x 3 x (8)²
= 0 + 1/2 x 3 x 8 x 8
∴ S = 96 m.
In other words, the motorboat covers a distance = S = 96 m Ans.
Q. 5. A driver of a car travelling at 52 kmh-1 applies the brakes and accelerates uniformly in opposite direction. The car stops in 5 s. Another driver going at 3 km h-¹ applies his brakes slowly and stops in 10 s. On the same graph paper plot the speed versus time graph for the two cars. Which of the two cars travelled farther after the brakes were applied ?
Sol.—In the figure AB and CD represent velocity-time graphs of two cars which have their speeds 52 kmh-¹ and 30 kmh-¹ respectively.
Distance covered by the first car before coming to rest
= Area between velocity – time graph and time axis
= Area of ΔAOB
= 1/2 × OB × AO
= 1/2 × 5 × {52 × 5/18}
= 25 × 52 / 36
= 325/9
= 36.1 m
Distance covered by second car before coming to rest = Area of Δ COD
= 1/2 × OD × OC
= 1/2 × 10 × {30 × 5/18)
= 5 × 30 × 5 / 18 = 41.7 m.
In this way, after applying brakes the second car would cover more distance than the first car.
Q. 6. Fig 1.11 shows the distance – time graphs of three objects A, B and C. Study the graph and answer the following questions :
(a) Which of the three is travelling the fastest ?
(b) Are all three ever at the same point on the road ?
(c) How far has C travelled when B passes A ?
(d) How far has B travelled by the time it passes C ?
Sol.— (a) A Velocity of A = Slope of PN
= 10 – 6 / 101 – 0
= 40/11 = 3.63 km h-1
Velocity of B = Slope of OM
= 6 – 0/0.7 – 0
= 60/7 = 8.57 km h-1
Velocity of C = Slope of QM
= 6 – 2/0.7 – 0
= 40/7 = 5.71 km h-1
Because slope of object B is maximum of all therefore, it is moving fastest.
(b) Since all the three graphs do not intersect at any point therefore, all the three do not meet ever at the same point on the road.
(c) When the object B passes A at point M (at 1.4 hr) then at that time the object C will be 9.3 km away from the origin O.
(d) B passes C after covering 8 km.
Q.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 it will strike the ground ? After what time will it strike the ground ?
Sol.— Here, u = 0 m s-1 [Dropped from rest]
S = u 20 m
a = 10 m s-2
v = ?
t = ?
Using, v2 – u2 = 2aS
v2 – (0)2 = 2 x 10 x 20
v2 = 400
v2 = √400
= √20 x 20
= 20 m s-1
Now, v = u + at
20 = 0 + 10 x t
or, t = 20/10
∴ t = 2s Ans.
Q. 8. Speed-time graph for a car is shown in the fig 1.13 :
(a) Find how far the car travelled in first 4 s. Shade the area on the graph that represents the distance travelled by car during this period.
(b) Which part of the graph represents uniform motion of the car ?
Sol.— (a) 5 small squares of x axis = 2 s
3 small squares of y axis = 2 m s-1
Area of 15 small squares = 2 s x 2 m s-1
= 4 m
∴ Area of 1 small squares = 4/15 m
Area of velocity-time graph under 0 to 5 s = 57 small squares +1/2 × 6 small squares.
= (57 + 3) small squares
= 60 small squares.
Distance covered by car in 4 s = 60 x 4/15 m
= 16 m Ans.
(b) After 6 s the car has uniform motion.
Q. 9. 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.
Ans.— (a) Yes, this situation is possible.
Example. When an object is projected upwards, its velocity at the maximum height is zero although acceleration on it is 9.8 m s-2 i.e. equal to g.
(b) Yes, at the maximum height of projection the velocity is in the horizontal direction and its acceleration is perpendicular to the direction of motion as shown in figure.
Q. 10. An artificial satellite is moving in a circular path orbit of radius 42,250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Sol.— Radius of circular path of artificial satellite (r) = 42,250 km
Angle formed (subtended) at the centre of earth (θ) = 2 π radian
Time taken by the satellite to complete 1 revolution (t) = 24 hrs
= 24 x 3600 s
= 86400 s
Angular velocity (w) = θ/t
= 2 π/86400 rad/s
Linear velocity of satellite (v) = r × w
= 42,250 x 2 π/86400 km/s
= 42250 x 2 x 3.14/ 86400 km/s
= 3.07 km/s Ans.
ADDITIONAL IMPORTANT QUESTIONS
LONG ANSWER TYPE QUESTIONS
Q. 1. Derive mathematically the equations of motion.
Ans.—Equations of motion under uniform acceleration. The following are the equations of motion of an object moving in a straight line with uniform acceleration.
Q. 2. For uniform accelerated motion, establish the following equations of motion by graphical method :
Ans.— Consider an object moving with initial velocity ‘u’ along a straight line and uniform acceleration ‘a’. Suppose its final velocity becomes v after time t and during this time it travels a distance S. If time is represented along X-axis and velocity along Y-axis, then velocity-time graph obtained will be BA, an inclined straight line. With its help we can derive the various equators of motion.
Here, OA = ED = u
OC = EB = v and OE = t = AD
(a) Equation for velocity – time relation. We know that
Acceleration = Slope of velocity-time graph AB
or, a = DB/AD
= DE/OE
= EB – ED/OE
= v – u / t [DE = OA = u]
or , v – u = at
or, v = u + at
This proves the first equation of motion.
(b) Equation for position-time relation
We know that area below velocity-time graph and time axis gives the distance covered by the body.
Distance travelled by the object in time t is
S = Area of the trapezium OABE
= Area of rectangle OADE +
Area of triangle ADB
Q. 3. Draw velocity-time graph for a body moving with uniform velocity. Hence show that the area under the velocity-time graph gives the distance travelled by the body in a given time interval.
Ans.— Distance covered as area under the velocity-time graph. In Fig. straight line PQ is the velocity-time graph of a body moving with a uniform velocity, ‘v’ represented by OP.
Area of rectangle ABCD
= AD x AB
= OP x AB
= v × (t₂-t₁)
= Velocity x time
= Distance travelled in
time interval (t₂-t₁)
Hence, the area under the velocity-time graph gives the distance travelled by the body in the given time interval.
SHORT ANSWER TYPE QUESTIONS
Q. 1. Define rest and motion. Give one example for each.
Ans.— Rest. A body is said to be at rest if it does not change its position with respect to its surroundings. A book lying on the table in a room is at rest with respect to the other objects in the room.
Motion. A body is said to be in motion if it changes its position with time with respect to its surroundings. A car running on the road is in motion with respect to the electricpole, trees along the roadside.
Q. 2. Show that rest and motion are relative terms.
Or
Can an object be at rest as well as in motion at the same time ?
Ans.— An object may be at rest relative to one surrounding object and at the same time it may be in motion with respect to some other object. For example, a passenger sitting in moving train is at rest relative to other co-passengers. But since he is sharing the motion of the train, so he is in motion relative to the outside trees, lamp posts etc. Thus, rest and motion are relative terms.
Q. 3. Give some points of differences between distance and displacement.
Or
Give two differences between distance and displacement.
Ans.— Differences between distance and displacement :
| Distance | Displacement |
| Distance is the length of the actual path covered by a body, irrespective of its direction of motion. | Displacement is the shortest distance between the initial and final positions of a body in a given direction. |
| Distance between two given points may be same or different for different paths chosen. | Displacement between two given points is always same. |
| It is a scalar quantity. | It is a vector quantity. |
| Distance covered is always positive or zero. |
Displacement covered may be positive, negative or zero.
|
Q. 4. What is meant by uniform motion? Give an example.
Ans.— Uniform motion. If an object covers equal distances in equal intervals of time, however small the time interval may be, then the motion of the object is said to be uniform motion. For example, suppose a bus moves 10 km in the first 15 minutes, 10 km in second 15 minutes, 10 km in third 15 minutes and so on. Then one can say that the bus is in uniform motion.
Q. 5. Define the term velocity. What is its SI unit? Is it a scalar or vector quantity ?
Ans.— Velocity. Velocity of a body is defined as the displacement produced per unit time. It is also defined as the speed of a body in a gien direction.
∴ Velocity = Displacement / Time
If ‘S’ is the distance travelled by a body in a given direction and t is the time taken to travel that distance, then the velocity ‘v’ is given by
v = S / t
The SI unit of velocity is m s -¹.
Velocity is a vector quantity because it requires both magnitude and direction for its complete description.
Q. 6. Explain the difference regarding initial velocity and nature of acceleration of six moving bodies as expressed by the following velocity-time graphs.
Ans.— (a) The body has uniform velocity and a = 0.
(b) The body has uniform acceleration and its initial velocity is zero.
(c) The body has some initial velocity and is under uniform retardation.
(d) The body has some initial velocity and uniform acceleration.
(e) The body has zero initial velocity and it has variable acceleration.
(f) The body is at rest from O to A, it has uniform acceleration from A to B, uniform velocity from B to C and from C to D, the body is under uniform retardation.
NUMERICAL PROBLEMS (SOLVED)
Q.1. A police car running on a highway with a speed of 30 km/h fires on the vehicle of thiefs running in the same direction at a speed of 192 km/h. If the velocity of the bullet is 150 m/s then with what velocity the bullet will hit the thiefs ?
Sol.— Velocity of the bullet fired = 150 m / s
= 150 x 18 / 5 km/h
= 540 km/h
∴ Total speed = Speed of police vehicle + Speed of bullet
= 30 km/h + 540 km/h
= 570 km/h
Speed of the bullet relative to the speed of thief’s vehicle in the same direction
= (570-192) km/h
= 378 km/h
= 378 × 1000 / 60 x 60 m/s
= 105 m/s Ans.
Q. 2. A train 50 m long travels on a plain and level track and reached a post in 5 secs. Find (i) speed of the train (ii) the time train will take to cross 450 m long bridge.
Sol.— (i) Since the train takes 5 secs to reach the post, it covers a distance equal to its own length, then
speed of train = Total distance travelled / Total time taken
= 50 m / 5 s
= 10 m/s
(ii) Distance travelled by the train to cross the bridge
= Length of the bridge + Length of train
= 450 m + 50 m
= 500 m
∴ Time taken to cross the bridge = Total distance covered / speed of the train
= 500 m / 10 m/s
= 50 s Ans.
Q. 3. A cheetah is the fastest land animal and can achieve a peak velocity of 100 km/h upto distances less than 500 m. If a cheetah spots his prey at a distance of 100 m. What is the minimum time it will take to get its prey, if the average velocity attained by it is 90 km/h.
Sol.— Here, v = 90 km/h
= 90 x 1000 m / 3600 s = 25 m/s
s = 100 m
Minimum time, t = s/v
= 100 / 25 = 4 s Ans.
Q. 4. A railway train 50 m long passes over a bridge 250 m long with uniform velocity of 10 m s -¹. How long will it take to completely pass over the bridge ?
Sol.— To completely pass over the bridge, the train will have to cover the length of the bridge, as well as its own length i.e., AB + BC
∴ Total distance to be covered, S= 50 + 250 = 300 m
Velocity of train, u= 10 m s -¹
Here velocity is uniform, hence
a = 0 m s –2
Using, S = ut + 1/2 at2
300 = 10t +0
or, t = 30 s
Hence train in will pass completely over the bridge in 30 s.
Q. 5. A car starts from rest and moves with uniform acceleration for 4 40 s. It then moves with uniform velocity attained by it and is finally brought to rest in 30 s under uniform retardation. The car covers a total distance of 1,380 m in 2.5 min. Calculate constant speed, acceleration and retardation.
Sol.— Let x be the constant speed. Now the total distance covered is equal to total area between velocity-time graph and time axis.
Therefore Distance covered = area of ΔABC + area of rectangle BCDE + area of ΔDEF
VERY SHORT ANSWER TYPE QUESTIONS
Q. 1. What is motion ?
Ans.— Motion. When an object changes its position with time then it is said to be in motion.
Q. 2. What is displacement of object ?
Ans.— Displacement. The change in position of an object in a given direction is known as displacement. It is measured by the shortest distance moved by an object from its initial position to final position.
Q. 3. Which device shows the speed of vehicles ?
Ans.— Odometer.
Q. 4. What is uniform motion ?
Ans.— Uniform motion. When a body covers equal distances in equal intervals of time then its motion is called uniform motion.
Q. 5. Give two examples of non-uniform motion.
Ans.— (i) A car moving on the road.
(ii) A man doing exercise in park.
Q. 6. Define speed.
Ans.— Speed. The distance travelled by a body in a unit time is called its speed.
Q. 7. What is the SI unit of speed ?
Ans.— S.I. unit of speed is m s -¹.
Q. 8. How is average speed obtained ?
Ans.— Average speed. It is the total distance travelled by an object divided by the time taken to travel Travel that distance.
Average speed = Total distance travelled / Total time taken
Q. 9. Define velocity ?
Ans.— Velocity. The speed in a particular direction is called velocity.
Or
It is the distance travelled by an object in a particular direction.
Q. 10. What is acceleration ?
Ans.— Acceleration. It is defined as the rate of change of velocity with time.
Acceleration = Change of velocity / Total time taken
Q. 11. What is the SI unit of acceleration ?
Ans— m s –2
Q. 12. A cricket player tosses the ball upward and again catches it. What is the total displacement?
Ans.— Total displacement is zero.
Q. 13. Is displacement a scalar or vector quantity ?
Ans.— Displacement is a vector quantity because it needs both magnitude and direction for its complete representation.
Q. 14. What would be acceleration of a body if its velocity-time graph is line parallel to the time axis ?
Ans.— Zero, as the body possesses uniform velocity.
Q. 15. A body is moving along the boundary of a square plot of land with constant speed. Does its velocity remain unchanged ?
Ans.— No. because its velocity changes due to change in direction.
Q. 16. What will be the position-time graph of a city bus standing at rest at a depot ?
Ans.— A straight line parallel to the time axis.
Q. 17. What is the nature of the distance time graph for an object moving uniformly along a straight long road ?
Ans.— A straight line inclined to the time axis.
Q. 18. Does the speedometer of a car measure its average speed ?
Ans.— No. It measures only instantaneous speed.
MULTIPLE CHOICE QUESTIONS
Choose the correct answer :
1. The SI unit of velocity is :
(A) m s –1
(B) m s –2
(C) m s –3
(D) Nm –1
Ans.— (A) m s –1
2. m/s² is the SI unit of :
(A) distance
(B) displacement
(C) velocity
(D) acceleration.
Ans.—(D) acceleration.
3. A car goes from town A to another town B with a speed of 40 km/h and returns back to the town A with a speed of 60 km/h. The average speed of the car during the complete journey is
(A) 48 km/h
(B) 50 km/h
(C) zero
(D) none of these.
Ans.— (A) 48 km/h
4. The initial velocity of a body is u. It is under uniform acceleration a. Its velocity v at any time t is given by :
(A) v = u + at²
(B) v = u+ 1/2 at²
(C) v = u + at
(D) v = u.
Ans.—(C) v = u + at
5. If the time-displacement graph of a particle is parallel to the time axis, the velocity of the particle is :
(A) unity
(B) zero
(C) infinity
(D) none of these.
Ans.— (B) zero
6. Ratio of angular velocity of earth to that of hour hand of a clock is :
(A) 2
(B) 1/2
(C) 1/12
(D) 1/24
Ans.— (B) 1/2
7. A body is moving with uniform velocity of 20 m s –1 for last 4 s, its acceleration is :
(A) 5 m s –2
(B) 0.2 m s –2
(C) zero
(D) 80 m s –1
Ans.— (C) zero
8. A body is moving with uniform speed along a circular path, its acceleration is :
(A) uniform
(B) variable
(C) zero
(D) negative
Ans.— (A) uniform
9. Speed is never :
(A) zero
(B) fraction
(C) negative
(D) positive
Ans.— (C) negative
10. The rate of change of displacement is :
(A) speed
(B) velocity
(C) acceleration
(D) retardation.
Ans.— (B) velocity
11. SI unit of displacment is :
(A) m
(B) m s –1
(C) m s –2
(D) none of these.
Ans.— (A) m
12. A particle covers equal distance in equal intervals of time, it is said to be moving As with uniform :
(A) speed
(B) velocity
(C) acceleration
(D) force
Ans.— (A) speed
13. Unit of angular velocity is :
(A) rad
(B) m s –1
(C) rad s –2
(D) rad s –1
Ans.— (D) rad s –1
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