AP 9th Class Physics 7th Lesson Important Questions Motion

These AP 9th Class Physics Important Questions 7th Lesson Motion will help students prepare well for the exams.

AP 9th Class Physics 7th Lesson Motion Important Questions

Class 9 Physical Science Chapter 7 Important Questions – 2 Marks

Question 1.
Define non – uniform acceleration.
Answer:
If a particle undergoes unequal changes of velocity in equal time intervals,, then it is in non – uniform velocity. Ex : Student riding a bicycle on a road.
Note : Variable acceleration results in thrills.

Question 2.
Define deceleration (or) retardation.
Answer:
If the velocity of a particle decreases with time is known as deceleration and the mo¬tion is known as decelerated (or) retarded motion.
Note : – ve of acceleration = deceleration.
Acceleration is measured in vehicles by accelerometer.

Question 3.
Why don’t we directly perceive the motion of the earth ?
Answer:

1. We don’t directly perceive the motion of the Earth because we are constantly moving with it.
2. Our senses are only able to detect changes in motion and since we are moving at a constant speed and direction, we do not sense any motion.

Question 4.
How do you measure the rate of motion of an object ? What does it mean ?
Answer:

1. One of the ways of measuring the rate of motion of an object is to find out the distance travelled by the object in unit time.
2. This quantity is referred to as speed.

Question 5.
An object moving with 6m per second execute an acceleration 2 m/s² in next 3 seconds. How much distance it covered ?
Solution:
u = 6 m/s; t = 3 sec; a = 2 m/s²
s = ut + \(\frac{1}{2}\) at² = 6 × 3 + \(\frac{1}{2}\) × 2 × 3² = 18 + 9 = 27 m
The object covers 27 m in 3 sec.

Question 6.
A car stopped after travelling distance 8 m due to applying brakes at the speed of 40 m/s. Find acceleration and retordation of car in that period.
Answer:
Here u = 40 m/s; v = 0 (vehical stopped); s = 8 m; a = ?
v² – u² = 2as
0 – 40² = 2 × a × 8 ⇒ a = \(\frac{-(40)(40)}{2 \times 8}\) = – 100 m/s²
Acceleration = 100 m/s² with retordation on (- sign).

Question 7.
Define acceleration and write its units.
Answer:

1. Acceleration is a measure of the change in the velocity of an object per unit time. That is, acceleration = change in velocity / time taken.
2. The S.I. unit of acceleration is ms^-2.

Question 8.
What is the acceleration of an object if the velocity of the object changes from an initial value u to the final value v in time t ?
Answer:
If the velocity of an object changes from an initial value u to the final value v in time t, the acceleration a is a = (v – u )/t.

Question 9.
What is meant by positive and negative acceleration ?
Answer:
The acceleration is taken to be positive if it is in the direction of velocity and negative when it is opposite to the direction of velocity.

Question 10.
When does acceleration say to be uniform? Give an example of uniform acceleration.
Answer:

1. If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration of the object is said to be uniform.
2. The motion of a freely falling body is an example of uniformly accelerated motion.

Question 11.
Explain non – uniform acceleration.
Answer:

1. In object can travel with non-uniform acceleration if its velocity changes at a nonuniform rate.
2. For example, if a car travelling along a straight road increases its speed by unequal amounts in equal intervals of time, then the car is said to be moving with non¬uniform acceleration.

Question 12.
Give an example when an object does not change its magnitude of velocity but only its direction of motion.
Answer:
An example of an object that does not change its magnitude of velocity but only its direction of motion is a satellite in orbit around the Earth. The satellite moves at a constant speed (magnitude of velocity) but changes direction as it orbits the Earth.
(or)
The motion of the athlete moving along a circular path is an example of an accelerated motion.

Question 13.
The motion of the athlete moving along a circular path is, therefore in uniform accelerated ipotion.
Find the velocity if the radius is r and time is t sec for one revolution.
Answer:

1. We know that the circumference of a circle of radius r is given by 2πr.
2. If the athlete takes t seconds to go once around the circular path of radius r, the speed v is given by v = \(\frac{2 \pi r}{t}\).

Question 14.
Define uniform motion ?
Answer:
If an object covers equal distances in equal time intervals, then the object is considered to be in uniform motion. Ex: Rotation of earth.

Question 15.
Define Non – Uniform motion.
Answer:
If an object covers unequal distances in equal time intervals, then the object is consid¬ered to be in non – uniform motion.
Ex: Person jogging in a park,
Car moving in a crowded street.

Question 16.
What is the average speed of an object ?
Answer:
Ratio of total distance travelled by total time taken.
S_Avg = \(\frac{\text { Total distance travelled }+ \text { Total time taken }}{2}\) ; V = \(\frac{s}{t}\)

Question 17.
What is average velocity ?
Answer:
Arithmetic mean of initial velocity and final velocity give average velocity.
Average velocity = \(\frac{\text { Initial velocity }+ \text { Final velocity }}{2}\)
Note : Velocity should be uniform.
V_Avg = \(\frac{\mathrm{u}+\mathrm{v}}{2}\) ; u = Initial velocity ; v = final velocity

Question 18.
Define acceleration.
Answer:
Change of velocity, of a particle per unit time is known as acceleration.
Acceleration = \(\frac{\text { change of velocity }+ \text { time }}{2}\) ⇒ a = \(\frac{v-u}{t}\)

Question 19.
Derive the S.I. unit of acceleration.
Answer:
Unit of velocity = ms^-1 ; Unit of time = s
Unit of acceleration a = \(\frac{v-u}{t}\) = \(\frac{\mathrm{ms}^{-1}}{\mathrm{~s}}\) = ms^-1 x s^-1 = ms^-2
Note : If a particles velocity changes with respect to time it is said to be in accelerated motion.

Question 20.
What is the average speed of a Cheetah that sprints 100m in 4 sec ? What if it sprints 50 m in 2 sec ?
Solution:
i) Distance = 100 m; Time = 4 sec
Average speed = \(\frac{100 \mathrm{~m}}{4 \mathrm{sec}}\) = 25 m/sec

ii) Distance = 50m ; Time = 2 sec
Average speed = \(\frac{50 \mathrm{~m}}{2 \mathrm{sec}}\) = 25 m/sec.

Question 21.
As shown in following figure, a point traverses the curved path.
Draw the displacement vector from given points A to B.

Answer:

As the point traverses from A to B, the displacement is the shortest distance between A and B. Hence the displacement vector will be as follows.

Question 22.
Explain the terms in the formula v = u + at.
Answer:
v = u + at
v = final velocity
u = initial velocity
a = acceleration
t = time

Motion Class 9 Important Questions – 3 marks

Question 1.
Two trains each of having a speed of 30 km/h are headed at each other in opposite direction on the same track. A bird flies off one train to another with a constant speed of 60 km/h when they are 60 km apart till before they crash. Find the distance covered by the bird and how many trips the bird can make from one train to the other before they crash.
Answer:
Speed of each train = 30 km/hr
Speed of the bird = 60 km/hr
Distance between the two trains = 60 km
These two trains crash in one hour.
The bird flies a distance of 60 km till before the two trains crash.
The bird can make number of trips (infinity) before they crash.

Question 2.
Sometimes we use average speed instead of the speed of an object. Why ?
Answer:

1. The speed of an object need not be constant.
2. In most cases, objects will be in non-uniform motion.
3. Therefore, we describe the rate of motion of such objects in terms of their average speed.

Question 3.
Explain the term average speed of an object with an example.
Answer:
The average speed of an object is obtained by dividing the total distance travelled by the total time taken.

1. That is, average speed = Total distance travelled /Total time taken.
2. If an object travels a distance ‘s’ in time ‘t’, then its speed ‘v’ is, v = s /t.
3. Let us understand this by an example.
4. A car travels a distance of 100 km in 2 h. Its average speed is 50 km h^-1.
5. The car might not have travelled at 50 km h^-1 all the time.
6. Sometimes it might have travelled faster and sometimes slower than this.

Question 4.
A car covers half the distance at a speed of 50 km/h and the other half at 40 km/h. Find the average speed of the car. (44.44 km/h).
Answer:
Let the total distance = x km.

Question 5.
Consider a train which can accelerate with an acceleration of 20 cm/s² and slow down with deceleration of 100 cm/s². Find the minimum time for the train to travel between the stations 2.7 km apart. (180 s)
Answer:
Let the Acceleration of the train α = 20 cm/s²
Deceleration of the train β = 100 cm/s²
Distance between the two stations s = 2.7 km = 27 × 10⁴ cm
Let the minimum time for the train to travel between the two stations is t sec.

Question 6.
What are the uses of graphs?

1. Graphs provide a convenient method to present basic information about a variety of events.
2. For example, in the telecast of a one-day cricket match, vertical bar graphs show the run rate of a team in each over.
3. In mathematics, a straight line graph helps in solving a linear equation having two variables.
4. To describe the motion of an object, we can use line graphs.

Question 7.
Draw a distance-time graph of an object moving with uniform speed.
Answer:

Question 8.
Interpret the given graphs.

Answer:

1. Fig. (a) shows a velocity-time graph that represents the motion of an object whose velocity is decreasing with time.
2. While Fig.(b) shows the velocity-time graph representing the non-uniform variation of velocity of the object with time.

Question 9.
What is the observation and conclusion from the activity which is shown in the figure ?

Answer:
Observation: On being released the stone moves along a straight line tangential to the circular path.

Reason : This is because once the stone is released, it continues to move along the direction it has been moving at that instant.

Conclusion : This shows that the direction of motion changed at every point when the stone was moving along the circular path.

Question 10.
What is uniform circular motion? Give three examples. Write the formula to calculate the velocity in a uniform circular motion.
Answer:
1) When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.
2) Examples :
a) There are many more familiar examples of objects moving under uniform circular motion, such as the motion of the moon and the earth, a satellite in a circular orbit around the earth, a cyclist on a circular track at a constant speed and so on.

b) The formula for calculating the velocity in a uniform circular motion is v = \(\frac{2 \pi r}{t}\)
(when radius r, speed v and time t is given).

Question 11.
How do you calculate the average velocity of an object, in case the velocity of the object is changing at a uniform rate ?
Answer:

1. In case the velocity of the object is changing at a uniform rate, then the average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time.
2. That is, average velocity = (initial velocity + final velocity)/ 2
3. Mathematically, v_ave =( u + v)/2 where v_ave is the average velocity, u is the initial velocity and v is the final velocity of the object.

Question 12.
What is the quantity which is measured by the area occupied below the velocity-time graph ?

Answer:
The quantity measured by the area occupied above the velocity-time graph is displacement or distance travelled.

Important Questions on Motion Class 9 – 5 marks

Question 1.

i) Calculate the speed from d – t graph between given points P & Q.
ii) Convert 126 kmh^-1 into ms^-1.
iii) Convert 60 ms^-1 into kmh^-1.
Answer:
i) 1) Draw ⊥ to time axis (X) from P.
2) Draw ⊥ to time axis (X) from Q.
3) From draw line ⊥ distance (Y – axis) meeting at R.
Ratio of \(\frac{\mathrm{QR}}{\mathrm{PR}}\) gives slope PQ.
Slope PQ denotes speed.
PQ = \(\frac{\mathrm{QR}}{\mathrm{PR}}\)

Question 2.
Define the following terms and give its S.I. units, a) distance b) displacement c) speed d) velocity e) acceleration.
Answer:
a) Distance is the total length covered by an object in motion. It is a scalar quantity, meaning it has magnitude but no direction. S.I. units for distance is metre (m).

b) Displacement is the change in position of an object with respect to a reference point or starting position. It is a vector quantity, meaning it has magnitude arid direction. S.I. units for displacement is metre (m).

c) Speed is the rate of change of distance with respect to time. It is a scalar quantity, meaning it has magnitude but no direction. S.I. units for speed is metre per second (m/s).

d) Velocity is the rate of change of displacement with respect to time. It is a vector quantity, meaning it has magnitude and direction. S.I. units for velocity is metre per second (m/s).

e) Acceleration is the rate of change of velocity with respect to time. It is a vector quantity, meaning it has magnitude and direction. S.I. units for acceleration is metre per second squared (m/s²).

Question 3.
How do you use distance-time graph to determine the speed of an object ?

Answer:

1. We can use the distance-time graph to determine the speed of an object.
2. To do so, consider a small part AB of the distance-time graph shown in given figure.
3. Draw a line parallel to the X – axis from point A and another line parallel to the Y – axis from point B.
4. These two lines meet each other at point C to form a triangle ABC. Now, on the graph, AC denotes the time interval (t₂ – t₁) while BC corresponds to the distance (s₂ – s₁).
5. We can see from the graph that as the object moves from the point A to B, it covers a distance (s₂ – s₁) in time (t₂ – t₁).
6. The speed, v of the object, therefore can be represented as v = \(\frac{s_2-s_1}{t_2-t_1}\).

Question 4.
Discuss whether your classroom walls are at rest or in motion.
Answer:

1. From the perspective of someone inside the classroom, the walls are at rest relative to them, as they are not moving in relation to their position.
2. However, from the perspective of someone outside the classroom, such as an observer watching the classroom move past them on a train, the walls are in motion relative to them.
3. The concept of relativity depends on the observer’s reference frame and there is no absolute state of rest or motion.
4. Therefore, whether the classroom walls are considered to be at rest or in motion depends on the observer’s reference frame.

Question 5.
Have you ever experienced that the train in which you are sitting appears to move while it is at rest ? Discuss and share your experience.
Answer:

1. Yes. I experienced the phenomenon.
2. This phenomenon is a result of our perception of motion being relative to our surroundings.
3. When a stationary train is located next to another moving train, the stationary train can experience a sensation of movement due to the movement of the other train.
4. This is because of the concept of relative motion, where our perception of motion is relative to our surroundings.
5. When the other train moves, it creates a visual and physical reference point for the stationary train, causing the observer to perceive their own train as moving.

Question 6.
Take a metre scale and a long rope. Walk from one corner of a basketball court to its opposite corner along its sides. Measure the distance covered by you and the magnitude of the displacement.
What difference would you notice between the two in this case ?
Answer:

1. Assume a standard basketball court size of 28 metres by 15 metres.
2. Distance covered : To walk from one corner to the opposite corner along the sides of the basketball court, you would need to cover a distance equal to one length + one breadth.
   = (28 + 15) = 43 metres
3. Displacement: Then, the diagonal distance would be the magnitude of the displacement.
   Diagonal distance = \(\sqrt{\left(28^2+15^2\right)}\) = \(\sqrt{(784+225)}\) = \(\sqrt{1009}\) = 31.77 metres.
4. Hence, distance covered is more than that of the displacement.

Question 7.
Automobiles are fitted with a device that shows the distance travelled. Such a device is known as an odometer. A car is driven from Bhubaneshwar to New Delhi. The difference between the final reading and the initial reading of the odometer is 1850 km. Find the magnitude of the displacement between Bhubaneshwar and New Delhi by using the Road Map of IndiAnswer:
Answer:

1. To find the magnitude of the displacement between Bhubaneshwar and New Delhi, we need to determine the straight-line distance between the two cities using the road map of IndiAnswer:
2. This displacement means the shortest distance between two points.
3. The straight-line distance between Bhubaneshwar and New Delhi is approximately 1165 km.
   This distance represents the magnitude of the displacement between the two cities.

Question 8.
At a time when it is cloudy, there may be frequent thunder and lightning. The sound of thunder takes some time to reach you after you see the lightning. Can you answer why this happens ? Measure this time interval using a digital wrist watch or a stop watch. Calculate the distance of the nearest point of lightning. (Speed of sound in air = 346 ms^-1) (Hint: Do yourself)
Answer:

• The reason why the sound of thunder takes some time to reach us after we see the lightning is because light travels much faster than sound. Light travels at the speed of light (approximately 299,792,458 metres per second).
• In contrast, sound travels much slower, at a speed of approximately 346 metres per second in air.
• To measure the time interval between seeing the lightning and hearing the thunder, you can use a stopwatch or a digital wristwatch with a timer function. When you see the lightning, start the timer and stop it when you hear the thunder. The difference between the two times will give you the time interval.
• The distance between us and the lightning can be calculated using the following
  formula:
  Distance = Speed × Time
• For example, if the time interval is 10 seconds, the distance between us and the lightning can be calculated as follows:
• Distance = 346 m/s × 10 s = 3460 metres
• Therefore, the nearest point of lightning is approximately 3460 metres away from us.

Question 9.
In your everyday life you come across a range of motions in which (a) acceleration is in the direction of motion, (b) acceleration is against the direction of motion, (c) acceleration is uniform, (d) acceleration is non-uniform. Can you identify one example each for the above type of motion ?
Answer:
Yes. I can identify one example each for the above types of motion :
a) Acceleration in the direction of motion : When a car is moving forward and the driver applies the accelerator, the car’s speed increases and the acceleration is in the direction of motion.

b) Acceleration is against the direction of motion: When a car is moving forward and the driver applies the brakes, the car’s speed decreases and the acceleration is against the direction of motion.

c) Uniform acceleration: A ball falling under the influence of gravity is an example of uniform acceleration. The acceleration remains constant throughout the motion, and the speed of the ball increases uniformly.

d) Non-uniform acceleration : A rocket taking off from the ground is an example of non-uniform acceleration. The rocket’s acceleration changes continuously during its ascent and its speed increases at varying rates.

Question 10.
Differentiate uniform motion and non-uniform motion.
Answer:

Feature	Uniform Motion	Non-Uniform Motion
Definition	Motion of an object with constant velocity	Motion of an object with changing velocity
Velocity	Constant	Changing
Acceleration	Zero	Non-zero
Example	A car moving on a straight highway at a constant speed	A car moving on a winding road or in city traffic where its speed keeps changing
Equation	d = vt	s = ut + 1/2 at2 (for motion with constant acceleration)

Question 11.
Differentiate speed and velocity.
Answer:

	Speed	Velocity
Definition	The distance travelled per unit time without regard to direction.	The rate at which an object changes its position with respect to a reference point, including both magnitude and direction.
Formula	Speed = distance/time	Velocity = displacement/time
Scalar or Vector	Scalar	Vector
S.J. Unit	Metres per second (m/s)	Metres per second (m/s)
Direction	No direction, only magnitude	1-las both magnitude and direction
Example	A car travels 100 km in 2 hours. The speed of the car is 50 km/h.	A car travels 100 km north in 2 hours. The velocity of the car is 50 km/h north.

Question 12.
Derive graphically velocity – time relation.
Answer:

1. Let a particle travelling with a certain initial velocity.
2. Initial velocity OA = CE = u at time t = 0.
3. The final velocity of the particle after reaching from A to B is v.
4. Final velocity OD = BE = v
5. Draw AC ⊥ BE.
6. From graph BC = BE – CE = v – u …………. (1)
7. From graph OE = AC = t ………….. (2)
   The slope of v -1 graph gives the acceleration of the particle from A to B.
   Slope AB = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
   Acceleration = \(\frac{\text { change of velocity }}{\text { change of time }}\) [by definition]
   
   This relation you can use when v, u, t & a are to be found.

Question 13.
Derive graphically position – time(or) displacement – time relation.
Answer:

1. Let a particle travelling with a certain initial velocity.
2. Initial velocity OA = CE = u at time t = 0.
3. The final velocity of the particle after reaching from A to B is v.
4. Final velocity OD = BE = v
5. Draw AC ⊥ BE.
6. From graph, BC = BE – CE = v – u …………. (1)
7. From graph, OE = AC = t …………. (2)
   The area under the velocity-time graph is equal to displacement (s).
   displacement = Ar of Quadrilateral OABE
   s = Area of rectangle OACE + Area of ΔABC
   s = l × b + \(\frac{1}{2}\) × b × h
   s = OE × OA + \(\frac{1}{2}\) × OE × BC [Base AC = OE]
   s = ut + \(\frac{1}{2}\)(v – u) × t ________ (1) ; v = u + at ; v – u = at ______ (2)
   Put (2) in (1)
   s = ut + \(\frac{1}{2}\) at × t ⇒ s = ut + \(\frac{1}{2}\)at²
   Note : This relation is used to find s, u, t and a.

Question 14.
Derive graphically position – velocity (or) velocity – displacement relation.
Answer:

1. Let a particle travelling with a certain initial velocity.
2. Initial velocity OA = CE = u at time t = 0.
3. The final velocity of the particle after reaching from A to B is v.
4. Final velocity OD = BE = v
5. Draw AC ⊥ BE.
6. From graph, BC = BE – CE = v – u …………… (1)
7. From graph, OE = AC = t …………… (2)
   The area of trapezium which is equivalent to displacement.
   Displacement = Area of trapezium OABE
   s = \(\frac{1}{2}\) (sum of parallel sides) × (distance between them)
   s = \(\frac{1}{2}\) (OA + BE). AC = \(\frac{1}{2}\) (v + u) .t ……….. (1) [∵ AC = OE = t]
   v = u + at ; v – u = at ; t = \(\frac{v-u}{a}\) …………… (2)
   Put (2) in (1)
   s = \(\frac{1}{2}\)(v + u)(\(\frac{v-u}{a}\))
   2as = (v + u) (v – u) [∵ (a + b) (a – b) = a² – b²]
   v² – u² = 2as
   Note : Use this to find v, u, s and a.

Question 15.
Describe uniform circular motion.
Answer:

When a particle moves in a circular path with uniform speed, its motion is called uniform circular motion.
2πr – length covered in one revolution
t = time revolution
v = \(\frac{2 \pi \mathrm{r}}{\mathrm{t}}\)
A particle moving along a circular path changes its direction continuously, its velocity is not constant, even if its speed is constant.
t = \(\frac{2 \pi \mathrm{r}}{\mathrm{v}}\) ; t = time period of revolution.
Note : Direction of a particle can be known be drawing a tangent at a point in-circular motion.

Extra Questions on Motion Class 9 – 4 marks

Question 1.

If a person is moved from point O to A and returned to C.
i) Calculate the distance and displacement of the person.
ii) If he moved from O to A and returned to A, then what is the displacement of the person ?
Answer:
i) Distance travelled by the person :
O to A = 60 km ; A to C = 35 km
Total distance travelled = 60 km + 35 km = 95 km
Displacement is 25 km.

ii) If the person moves from O to A and back to A, then the displacement of the person would be zero, because the person ends up at the same point where he started (i.e., point O). The person travels a distance of 120 km (60 km from O to A and back), but the displacement is zero as the person’s final position is the same as the initial position.

Question 2.
Below table shows the distance travelled by a car in a time interval of two seconds. Plot the distance-time graph for accelerated motion. Comment the shape of the graph.

Time in seconds	Distance in metres
0	0
2	1
4	4
6	9
8	16
i0 ,	25
12	36

Answer:

The nature of this graph shows non-linear variation of the distance travelled by car with time. Thus, the graph shown in below figure represents motion with non-uniform speed.

Question 3.
How do you calculate the displacement from the given graph ?

Answer:

1. We know that the product of velocity and time gives displacement of an object moving with uniform velocity.
2. The area enclosed by velocity-time graph and the time axis will be equal to the magnitude of the displacement.
3. To know the distance moved by the car between time t₁ and t₂ using above figure,
   draw perpendiculars from the points corresponding to the time t₁ and t₂ on the graph.
4. The velocity of 40 km h^-1 is represented by the height AC or BD and the time (t₂ – t₁) is represented by the length AB.
5. So, the distance ‘s’ moved by the car in time (t₂ – t₁) can be expressed as
   s = AC × CD
   = [(40 kmh^-1) × (t₂ – t₁)h]
   = 40 (t₂ – t₁) km
   = area of the rectangle ABDC (shaded in Fig.).

Question 4.
How do you determine the distance moved by the car from its velocity-time graph as shown in the figure (graph) ?

Answer:

1. We can determine the distance moved by the car from its velocity-time graph.
2. The area under the velocity-time graph gives the distance (magnitude of displacement) moved by the car in a given interval of time.
3. If the car would have been moving with uniform velocity, the distance travelled by it would be represented by the area ABCD under the graph.
4. Since the magnitude of the velocity of the car is changing due to acceleration, the distance ‘s’ travelled by the car will be given by the area ABCDE under the velocitytime graph.
5. That is, s = area of ABCDE
   = area of the rectangle ABCD + area of the triangle
   ADE = AB × BC + 1/2 (AD × DE)

Question 5.
The times of arrival and departure of a train at three stations A, B and C and the distance of stations B and C from station A are given in table. Plot and interpret the distance-time graph for the train assuming that its motion between any two stations is uniform.

Station	Distance from A (km)	Time of arrival (hours)	Time of departure (hours)
A	0	08.00	08:15
B	120	11:15	11:30
C	180	13.00	13:15

Answer:

The distance-time graph for the train shows that the train travels at a constant speed between each pair of stations, represented by the straight lines on the graph. The slope of each line represents the speed of the train, which is distance divided by time.

From the graph, we can also see that the train travels a total distance of 180 km in 4.5 hours, which gives aij average speed of 40 km/h.

Based on the given information answer the following questions.

Question 6.
The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of trip. If the trip took 8h, then …………….
1) What is the distance travelled by the vehicle in 8 hrs ?
A) 2400 km
B) 400 km
C) 200 km
D) 2000 km
Answer:
B) 400 km

2) What is the average speed ?
A) 50 kmh^-1
B) 50 ms^-1
C) 100 cms^-1
D) None
Answer:
A) 50 kmh^-1

3) What is the average speed in ms^-1 ?
A) 50 ms^-1
B) 22 ms^-1
C) 13.9 ms^-1
D) 15.4 ms^-1
Answer:
C) 13.9 ms^-1

4) With what fraction we have to multiply to convert kmh^-1 to ms^-1 ?
A) \(\frac{2}{3}\)
B) \(\frac{5}{6}\)
C) \(\frac{8}{7}\)
D) \(\frac{5}{18}\)
Answer:
D) \(\frac{5}{18}\)

Question 7.
Pragna swims 90 m long pool. She covers 180 m in one minute. By swimming from one end to the other.
Based on the above information answer the following questions.
1) Distance covered by Pragna in 60 sec.
A) 90 m
B) 0 m
C) 180 m
D) -90 m
Answer:
C) 180 m

2) Displacement of Pragna in 1 min
A) 180 m
B) 90 m
C) 0 m
D) -180 m
Answer:
C) 0 m

3) Average speed of Pragna
A) 3 ms^-1
B) 2 ms^-1
C) 12 ms^-1
D) None
Answer:
A) 3 ms^-1

4) Average velocity of Pragna
A) 3 ms^-1
B) -3 ms^-1
C) 2 ms^-1
D) 0 ms^-1
Answer:
D) 0 ms^-1

Question 8.
Sanjay paddles his cycle from his (i) Home to school attain a velocity of 6 ms^-1 in 30 sec. (ii) Then he applies brakes and the bicycle velocity came down to 4 ms^-1 in next 5 sec. Based on the above information, answer the following questions.
1) What is the initial velocity of Sanjay in first case ?
A) 6 ms^-1
B) 4 ms^-1
C) 2 ms^-1
D) 0 ms^-1
Answer:
A) 6 ms^-1

2) What is the acceleration of the bicycle in first case ?
A) 0.2 ms^-1
B) 0.2 ms^-2
C) 6 ms^-2
D) All
Answer:
B) 0.2 ms^-2

3) What is the initial velocity before brakes were applied ?
A) 4 ms^-1
B) 6 ms^-1
C) 0 ms^-1
D) None
Answer:
B) 6 ms^-1

4) What is the acceleration in case (ii) ?
A) 4 ms^-2
B) 0.4 ms^-2
C) -0.4 ms^-1
D) 2 ms^-2
Answer:
C) -0.4 ms^-1

Question 9.
Study the graph and answer the questions below.
The velocity of a car at regular intervals of time given below.

Time (s)	Velocity of Car (ms-1)
0	0
5	2.5
10	5.0
15	7.5
20	10.0
25	12.5
30	15.0

A graph is given below plotting the above details. Study the graph and answer the following questions.

1) What is the velocity of car at A ?
Answer:
5.0 ms^-1

2) What is the velocity of car at B ?
Answer:
10.0 ms^-1

3) What is the acceleration of car between A & B ?
Answer:
0.5 ms^-2

4) What is the displacement of car between A & B ?
Answer:
50 m

9. Study the graph and answer the questions below.
The velocity of a car at regular intervals of time given below.

Question 10.
Study the speed – time graph given below and answer the questions given below. Identify the speed of the particle by observing s – t graph.
i)
Answer:
Particle is moving with constant speed.
Reason : s – t graph is straight line parallel to the X – axis indicates particle with constant speed.

ii)
Answer:
Particles speed is increasing at constant rate.
Reason : Speed of particle increasing an uniformly.

iii)
Answer:
Particles speed is decreasing at a constant rate.
Reason : Speed is becoming zero at t = 10 sec.

iv)
Answer:
Speed increases from 0 to 5 sec constantly. Particle is travelling at constant speed from 5 to 10 sec and then speed decreases constantly from 10 to 15 sec.

Question 11.
The Area under the speed-time graph denotes which physical quantity ?
Answer:
Distance s = \(\frac{1}{2}\) × b × h
s = \(\frac{1}{2}\) × OA × BA

Question 12.
Identify which bike is moving with greater acceleration from the graph.

Answer:
Bike B is moving with greater acceleration.
Reason : The greater the slope the greater the acceleration so, bike B has more slope.

Question 13.
Identify the acceleration from the velocity-time graph given below.

Answer:
The particle is undergoing non – uniform acceleration.
Reason : v – t graph is not linear (straight – line).