Haryana State Board HBSE 9th Class Science Solutions Chapter 8 Motion Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Science Solutions Chapter 8 Motion

**HBSE 9th Class Science Motion Intext Questions and Answers**

Question 1.

An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer:

Yes, displacement can be zero, if an object covers some distance, e.g., let an object moves from origin point along a straight line to point ‘A’ by covering a distance of 55 km and covers the same distance of 55 km from ‘A’ to ‘O’ on return as shown in figure In this state-

Distance travelled by the object = 55 km + 55 km = 110 km

Displacement = Zero (0)

Question 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?

Solution:

Side of the square field ABCD = 10 metre

The perimeter of the square field ABCD = 4 × side

= 4 × 10 = 40 metres

Time taken for one round around the boundary = 40 s

Total time = 2 minutes 20 s (2 × 60 + 20) s = 140 s

Now, the distance covered by the farmer in 40 s = 40 m

The distance covered by the farmer in 1s m = \(\frac {40}{40}\) = 1 m

The distance covered by the farmer in 140 s 1 × 140 = 140 metres

If, the farmer starts moving from origin point A, then he will be at point C after covering a distance of 140 metres. In this way, the displacement of the farmer will be AC (the diagonal of the square.).

∴ AC = \(\sqrt{(\mathrm{AB})^2+(\mathrm{BC})^2}\)

= \(\sqrt{(10)^2+(10)^2}\)

= \(\sqrt{100+100}\)

= \(\sqrt{200}\)

= \(\sqrt{2 \times 100}\)

= 10\(\sqrt{2}\) m

= 10 × 1.414 = 14.14 m

= 10 × 1.414 = 14.14 m

= 10 × 1.414 = 14.14 m

= 10 × 1.414 = 14.14 m

= 10 × 1.414 = 14.14 m

Question 3.

Which of the following is true for displacement ?

(a) It cannot he zero.

(b) Its magnitude is greater than the distance travelled by the object.

Answer:

Both (a) and (b) are wrong for displacement.

Questions from Sub-section 8.2

Question 1.

Distinguish between speed and velocity.

Answer:

The differences between speed and velocity are as follows:

Speed:

1. It is the distance travelled by the body in unit time interval in any direction.

2. It is a scalar quantity, which has only magnitude.

3. It is always positive.

Velocity

1. It is the rate of distance travelled by the body in unit time interval in specified direction.

2. It is a vector quantity, which has both magnitude as well as direction.

3. It can be positive and negative.

Question 2.

Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer:

If the object is moving with uniform motion in a definite direction, then the magnitude of the average velocity of that object will be equal to its average speed.

Question 3.

What does the odometer of an automobile measure ?

Answer:

The odometer of the automobile measures the distance covered by the automobile.

Question 4.

What does the path of an object look like when it is in uniform motion ?

Answer:

When the object is in uniform motion, then its path seems as a straight line.

Question 5.

During an experiment, a signal from a spaceship tTactretf the ground-statianju five minutes. What was the distance of the spaceship from the ground station ? The signal travels at the speed of light, that is, 3 × 10^{8} ms^{-1}.

Solution:

The time taken by the signal from spaceship to earth = 5 minutes = 5 x 60 = 300 s

Speed of signal = 3 × 10^{8}ms^{-1}

∴ The distance of spaceship from ground station = Speed × Time = 3 × 10s × 300m = 9 × 1010m

Questions from Sub-section 8.3

Question 1.

When will you say a body is in (i) uniform acceleration (ii) non-uniform acceleration ?

Ans.

(i) Uniform acceleration : A body is said to be moving with uniform acceleration if there is equal change in the velocity of the body in equal interval of time.

(ii) Non-uniform acceleration : A body is said to be moving with non-uniform acceleration, if its velocity changes unequally.

Question 2.

A bus decreases its speed from 80 kmh^{-1} to 60 kmh^{-1} in 5 s. Find the acceleration of the bus.

Solution:

Initial speed of the bus (u) = 80 kmh^{-1}

= \(\frac{80 \times 1000}{3600}\)ms^{-1} = \(\frac{200}{9}\)ms^{-1}

Final speed of the bus (v) = 60 kmh^{-1}

\(\frac{60 \times 1000}{3600}\)ms^{-1}

\(\frac{50}{3}\)ms^{-1}

Time (t) = Ss

Acceleration of the bus (a) = \(\frac{v-u}{t}\) = \(\frac{\left(\frac{50}{3}-\frac{200}{9}\right) \mathrm{ms}^{-1}}{5 \mathrm{~s}}\) = \(\frac{150-200}{9 \times 5}\)ms^{-2}

\(\frac{-50}{45}=\frac{-10}{9}\)ms^{-2} = -1.1 ms^{-2}

Question 3.

A train starting from a railway station and moving with uniform acceleration attains a speed of 40 kmh1 in 10 minutes. Find its acceleration.

Solution:

The initial speed of the train (u) = 0

The final speed of the train (v) = 40 kmh^{-1}

= \(\frac{40 \times 1000}{3600}\) = \(\frac{40 \times 1000}{3600}\)

Time (t) = 10 minutes = 10 x 60s = 600s

∴ Acceleration (a) = \(\frac{v-u}{t}\) = \(\frac{\frac{100}{9}-0}{600}\)ms^{-1}

= \(\frac{100}{9 \times 600}=\frac{1}{54}\) ~ 0.02 ms^{-2}

Questions from Sub-section 8.4

Question 1.

What is the nature of the distance-time graps for uniform and non-uniform motion of an object?

Ans.

For a uniform motion, the graph of the distance covered with time is a straight line. In graph, part OB shows that the distance is increasing with uniform rate.

Distance-time graph for an object possessing non-uniform motion is a curved line as shown in the graph. .

Question 2.

What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Answer:

If the distance-time graph of an object, parallel to time axis is a straight line, then the object will move with uniform speed.

Question 3.

What can you say about the motion of an object, if its speed-time graph is a straight line parallel to the time axis ?

Answer:

If the speed-time graph of an object, parallel to time axis is a straight line, then the object will move with uniform speed.

Question 4.

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

Answer:

The quantity measured by the area occupied below the velocity-time graph shows the distance covered by the body.

Questions front Sub-section 8.5

Question 1.

Abus starting from rest moves with a uniform acceleration of 0.1ms-2for 2 minutes. Find (a) the speed acquired (b) the distance travelled.

Solution:

Here,

Initial speed of the bus (u) = 0

Final speed of the bus (v) = ?

Acceleration (a) = 0.1 ms

Time (t) = 2 minutes = 2 x 60s = 120 s

(a) The final speed (v) of the bus = u + at = 0 + (120) + \(\frac {1}{2}\) = 12 ms^{-1}

(b) The distance travelled by the bus, (s) = ut + \(\frac {1}{2}\) at^{2}

= 0(120) + \(\frac {1}{2}\)(0.1)(120)^{2}

Question 2.

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 go before it is brought to rest?

Solution:

Here,

Initial speed of the train (u) = 90 km h^{-1}

\(\frac{90 \times 1000}{3600}\) ms^{-1}= 25 ms^{-1}

Final speed of the train (v) = 0

Acceleration (a) = – 0.5 ms^{2}

Distance (s) = ?

We know that, v^{2} – u^{2} = 2as

s = \(\frac{90 \times 1000}{3600}\)

= \(\frac{(0)^2-(25)^2}{2(-0.5)}=\frac{-625}{-1.0}\) = 625 metres

∴ The train will cover a distance of 625 metres, before it is brought to rest.

Question 3.

A trolley, while going down on inclined plane, has an acceleration of 2 ms^{-2}. What will be is velocity 3s after the start?

Solution:

Here,

The initial speed of the trolley (u) = 0

Final speed of the trolley (v) = ?

Acceleration (a) = 2 ms^{-2}

Time (t) = 3s

We know that, v = u + at

= 0 + 2(3) = 6 ms^{-1}

∴ The velocity of the trolley after 3 s will be 6 ms^{-1}.

Question 4.

A racing car has a uniform acceleration of 4ms2. What distance will it cover in 10 s after start?

Solution:

Here,

The initial speed of the car (u) = 0

Acceleration (a) = 4 ms

Time (t) = 10s

Distance (s) = ?

We know that,

s = ut + \(\frac {1}{2}\) at^{2}

= 0 x (10) + \(\frac {1}{2}\) (4) (10)^{2}

= \(\frac {1}{2}\) × 4 × 100 = 200m

So, the covered distance will be 200m

Question 5.

A stone is thrown in a vertically upward direction with a velocity of 5 ms. If the acceleration of the stone during its motion is 10 ms2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Solution:

Here,

The initial velocity of the stone (u) = 5 ms^{-1}

Final velocity of the stone (v) = 0

Acceleration (a) = – 10 ms^{-2}

Maximum height (s) = ?

We know that,

v^{2}– u^{2} = 2as

⇒ S = \(\frac{v^2-u^2}{2 a}=\frac{(0)^2-(5)^2}{2(-10)}\) = \(\frac {-25}{20}\) = 1.25m

Applying, v = u + at

⇒ t = \(\frac{v-u}{a}=\frac{0-5}{-10}\) = 0.5 s

**HBSE 9th Class Science Motion Textbook Questions and Answers**

Question 1.

An athlete completes one round of a circular track of diameter 200 metre in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Solution:

Here,

The diameter of the circular track = 200 m

Radius (r) of the circular track = \(\frac {200}{2}\) = 1oo m

The circumference of the circular track = 2πr

= 2 x \(\frac {22}{7}\) × 1oo m

= \(\frac {4400}{7}\)m

Total time = 2 minutes 20 se

= (2 x 60 + 20)s (120 + 20)s = 140s

∴ The distance covered by the athlete in 40 s = \(\frac {4400}{7}\) m

The distance covered by the athlete in 1 s = \(\frac{4400}{7 \times 40}\) m

The distance covered by the athlete in 140 s = \(\frac{4400}{7 \times 40}\) x 140m = 2200 metres

Here, the total time is 140 s, in which the athlete will complete three complete rounds and one half round. If he will start from point A of the circumference and reach at point B.

In this way, displacement (AB) = The diameter of the circular track = 200 metres

Question 2.

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 s and then returns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speed and velocitie in jogging (a) from point A to B (b) from point A to C?

Solution:

(a) The total distance covered from point A to B = 300 m

The total time taken from point A to B = 2 min 30 s = (2 x 60 + 30)s = 150 s

Total displacement from point A to B = 300 – 0 = 300 m

= \(\frac {4400}{7}\) = 2 ms

(b) Total distance covered from point A to C = AB + BC = (300 + 100)m = 400 m

Total time taken fr om point A to C = 2 minutes 30 s + 1 minute

= (2 x 60 + 30)s + 60s = (150 + 60)s = 210s

Total displacement from point A to C = 300 – 100 = 200 m

= \(\frac {400}{210}\) = 1.90 ms^{-1}

= \(\frac {200}{210}\) = 0.952 ms^{-1}

= 0.952 ms^{-1}

Question 3.

Abdul, while driving to school, computes the average speed for his trip to be 20km h^{-1} on his return trip along the same route there is less traffic and the average speed is 30km h^{-1} . What is the average speed for Abdul’s trip?

Solution:

Average speed during the school trip v_{1} = 20 km h^{-1}

Average speed for return trip v_{2} = 30 km h^{-1}

Let, the distance from home to school = x km

Time taken for going school Speed = =\(\frac {x}{20}\) h

Time taken to return from school = =\(\frac {x}{30}\) h

Total distance covered for both (going and returning) trips = x + x = 2x km

Total time taken = \(\left(\frac{x}{20}+\frac{x}{30}\right) h\) = \(\left(\frac{3 x+2 x}{60}\right) h=\frac{5 x}{60} h=\frac{x}{12} h\)

∴ Average speed of Abdul during whole trip = = \(\frac{\frac{2 x}{x}}{12}=\frac{2 x \times 12}{x}=24 \mathrm{kmh}^{-1}\)

Question 4.

A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3 mr^{2} for 8 seconds. How far does the boat travel during this time?

Solution:

Here,

Initial speed of motorboat on lake (u) = 0

Acceleration (a) = 3.0 m^{-2}

Time (t) = 8.0 s

Distance (s) = ?

We know that, s = ut + at^{2}

= 0(8) + (3) (8)^{2}

= \(\frac {1}{2}\) × 3 × 64 = 96m

∴ Motorboat will cover a distance of 96m on lake. Ans.

Question 5.

A driver of a car travelling at 52km h applies the brakes and accelerates uniformly in the opposite direction. The car stops in Ss. Another driver going at 30km h in another car applies his brakes slowly and stops in lOs. On the same graph paper, plot the speed versus time graph for two cars. Which of the two cars travelled farther after the brakes were applied?

Solution:

Speed-time graph for both drivers is shown in the figure. Suppose.

first driver starts from point A and second driver starts from point B.

Distance covered by first car before rest = area of ∆ AOC

= \(\frac {1}{2}\) × base × height

\(\frac {1}{2}\) × 5s × 52km/h

\(\frac {1}{2}\) × 3600 × 52 km

=\(\frac {1}{2}\) × 300 × 52 x 1000 m

= 36.11 metres

In same way, the distance covered by second car before

rest = area of ∆BOD

= \(\frac {1}{2}\) × base × height

= \(\frac {1}{2}\) × 10 s × 30 km/h

= \(\frac {1}{2}\) × \(\frac {10}{3600}\) × 30km

= \(\frac {1}{2}\) × \(\frac {10}{3600}\) × 30 × 1000 m = 41.67m

It is clear from above solution that second car travels larger distance after applying brakes.

Question 6.

Figure shows the distance-time graph 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?

(e) How far has C travelled whenB passes A?

(d) How far has B travelled by the time it passes C?

Solution:

(a) B is travelling fastest as the slope in the graph of B is maximum as compared to A and C.

(b) Three can never be at the same point on the road because the three graphs do not meet at any single point.

(c) C has travelled the distance of 9 km, when B passes A.

(d) B has travelled a distance more than 4 km during the time he passed C.

Question 7.

A ball is gently dropped from a height of 20m. If its velocity increases uniformly at the rate of 10ms2, with what velocity will it strike the ground ? After what time will it strike the ground?

Solution:

Here,

Initial velocity of ball (u) = 0

Height (s) = 20 m

Acceleration (a) = 10 ms^{-2}

Final velocity of ball (v) = ?

Time (t) = ?

We know that,

v^{2} – u^{2} = 2as

v^{2} = u^{2} + 2as

= (0)^{2} + 2(10)(20) = 400

or v = \(\sqrt{400}\) = 20 ms^{-1}

Now, acceleration, a = \(\frac{v-u}{t}\)

or t = \(\frac{v-u}{a}\) = \(\frac{20-0}{10}\) = 2s

∴The velocity of the ball will be 20 ms’ before it strike the ground and it will strike the ground in 2s.

Question 8.

The speed-time graph for a car is shown in figure.

(a) Find how far does the car travel in first 4 seconds? Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

Solution:

(a) The area of the distance covered by the car in first 4 s is OAB, that is almost a right angled triangle.

∴The distance covered by the car in first 4s = \(\frac {1}{2}\) × OA × AB

The distance covered by the car in first 4s = \(\frac {1}{2}\) × 4 × 6 = 12m

The distance covered by the car OABin the figure.

(b) In the graph, the speed after 6 s, shows the uniform motion of the car.

Question 9.

State which of the following situations are possible and give an eamnle for eich 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.

Solution:

(a) Yes, this situation is possible. When a body is thrown up with some velocity, velocity is zero at the highest point but acceleration is non-zero and constant.

(b) Yes, this is possible. A body moving with uniform velocity on circular path is its example.

Question 10.

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.

Solution:

Here,

Radius of circular orbit of artificial satellite (r) = 42250 km

‘Time taken by the satellite to revolve around the earth (t) = 24 hours

= 24 × 3600s = 86400s

Speed of artificial satellite (v) = \(\frac {2πr}{t}\) = \(\frac{2 \times 3.14 \times 42250}{86400}\) k ms^{-1} = 3.07 k ms^{-1}

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