Science for Class 9 Science Chapter 8 - Motion

Answer:

Yes. An object that has moved through a distance can have zero displacement. Displacement is the shortest measurable distance between the initial and the final position of an object. An object which has covered a distance can have zero displacement, if it comes back to its starting point, i.e., the initial position.

Consider the following situation. A man is walking in a square park of length 20 m (as shown in the following figure). He starts walking from point A and after moving along all the corners of the park (point B, C, D), he again comes back to the same point, i.e., A.

In this case, the total distance covered by the man is 20 m + 20 m + 20 m

Answer:

The farmer takes 40 s to cover 4 × 10 = 40 m.

In 2 min and 20 s (140 s), he will cover a distance

Therefore, the farmer completes rounds (3 complete rounds and a half round) of the field in 2 min and 20 s.

That means, after 2 min 20 s, the farmer will be at the opposite end of the starting point.

Now, there can be two extreme cases.

Case I: Starting point is a corner point of the field.

In this case, the farmer will be at the diagonally opposite corner of the field after 2 min 20 s.

Therefore, the displacement will be equal to the diagonal of the field.

Hence, the displacement will be

Case II: Starting point is the middle point of any side of the field.

In this case the farmer will be at the middle point of the opposite side of the field after 2 min 20 s.

Therefore, the displacement will be equal to the side of the field, i.e., 10 m.
 

For any other starting point, the displacement will be between 14.1 m and 10 m.

Answer:

(a) Not true

Displacement can become zero when the initial and final position of the object is the same.

(b) Not true

Displacement is the shortest measurable distance between the initial and final positions of an object. It cannot be greater than the magnitude of the distance travelled by an object. However, sometimes, it may be equal to the distance travelled by the object.

 

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,

Hence, the speed of the artificial satellite is 3.069 km/s.

Answer:

Answer:

If the total distance covered by an object is the same as its displacement, then its average speed would be equal to its average velocity.


 

Answer:

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

 

Answer:

An object having uniform motion has a straight line path.

 

Answer:

Time taken by the signal to reach the ground station from the spaceship

= 5 min = 5 × 60 = 300 s

Speed of the signal = 3 × 10⁸ m/s

∴Distance travelled = Speed × Time taken = 3 × 10⁸ × 300 = 9 × 10¹⁰ m

Hence, the distance of the spaceship from the ground station is 9 × 10¹⁰ m.


 

Answer:

(i) A body is said to have uniform acceleration if it travels in a straight path in such a way that its velocity changes at a uniform rate, i.e., the velocity of a body increases or decreases by equal amounts in an equal interval of time.

(ii) A body is said to have non-uniform acceleration if it travels in a straight path in such a way that its velocity changes at a non-uniform rate, i.e., the velocity of a body increases or decreases in unequal amounts in an equal interval of time.

 

Answer:

Initial speed of the bus, u = 80 km/h

Final speed of the bus, v = 60 km/h

Time take to decrease the speed, t = 5 s

Here, the negative sign of acceleration indicates that the velocity of the car is decreasing.

 

Answer:

Initial velocity of the train, u = 0 (since the train is initially at rest)

Final velocity of the train, v = 40 km/h

Time taken, t = 10 min = 10 × 60 = 600 s

Hence, the acceleration of the train is 0.0185 m/s².

 

Answer:

The distance−time graph for uniform motion of an object is a straight line (as shown in the following figure).

The distance−time graph for non-uniform motion of an object is a curved line (as shown in the given figure).

Answer:

When an object is at rest, its distance−time graph is a straight line parallel to the time axis.

A straight line parallel to the x-axis in a distance−-time graph indicates that with a change in time, there is no change in the position of the object. Thus, the object is at rest.

 

Answer:

Object is moving uniformly.

A straight line parallel to the time axis in a speed−time graph indicates that with a change in time, there is no change in the speed of the object. This indicates the uniform motion of the object.


 

Answer:

Distance

The graph shows the velocity−time graph of a uniformly moving body.

Let the velocity of the body at time (t) be v.

Area of the shaded region = length × breath

Where,

Length = t

Breath = v

Area = vt = velocity × time …(i)

We know,

$Velocity = \frac{Displacement}{TIme }$

∴ Displacement = Velocity × Time…(ii)

From equations (i) and (ii),

Area = Displacement

Hence, the area occupied below the velocity−time graph measures the displacement covered by the body.

 

Answer:

(a) 12 m/s (b) 720 m

(a) Initial speed of the bus, u = 0 (since the bus is initially at rest)

Acceleration, a = 0.1 m/s²

Time taken, t = 2 minutes = 120 s

Let v be the final speed acquired by the bus.

∴v = 12 m/s

(b) According to the third equation of motion:

v² − u² = 2as

Where, s is the distance covered by the bus

(12)² − (0)² = 2(0.1) s

s = 720 m

Speed acquired by the bus is 12 m/s.

Distance travelled by the bus is 720 m.


 

Answer:

Initial speed of the train, u = 90 km/h = 25 m/s

Final speed of the train, v = 0 (finally the train comes to rest)

Acceleration = −0.5 m s⁻²

According to third equation of motion:

v² = u² + 2 as

(0)² = (25)² + 2 (−0.5) s

Where, s is the distance covered by the train

The train will cover a distance of 625 m before it comes to rest.

Answer:

Initial velocity of the trolley, u = 0 (since the trolley was initially at rest)

Acceleration, a = 2 cm s⁻² = 0.02 m/s²

Time, t = 3 s

According to the first equation of motion:

v = u + at

Where, v is the velocity of the trolley after 3 s from start

v = 0 + 0.02 × 3 = 0.06 m/s

Hence, the velocity of the trolley after 3 s from start is 0.06 m/s.

 

Answer:

Initial velocity of the racing car, u = 0 (since the racing car is initially at rest)

Acceleration, a = 4 m/s²

Time taken, t = 10 s

According to the second equation of motion:

Where, s is the distance covered by the racing car

Hence, the distance covered by the racing car after 10 s from start is 200 m.


 

Answer:

Initially, velocity of the stone,u = 5 m/s

Final velocity, v = 0 (since the stone comes to rest when it reaches its maximum height)

Acceleration of the stone, a = acceleration due to gravity, g = 10 m/s²

(in downward direction)

There will be a change in the sign of acceleration because the stone is being thrown upwards.

Acceleration, a = −10 m/s²

Let s be the maximum height attained by the stone in time t.

According to the first equation of motion:

v = u + at

0 = 5 + (−10) t

According to the third equation of motion:

v² = u² + 2 as

(0)² = (5)² + 2(−10) s

Hence, the stone attains a height of 1.25 m in 0.5 s.

 

Answer:

Diameter of a circular track, d = 200 m

Radius of the track, r =

Circumference = 2πr = 2π (100) = 200π m

In 40 s, the given athlete covers a distance of 200π m.

In 1 s, the given athlete covers a distance =

The athlete runs for 2 minutes 20 s = 140 s

∴Total distance covered in

The athlete covers one round of the circular track in 40 s. This means that after every 40 s, the athlete comes back to his original position. Hence, in 140 s he had completed 3 rounds of the circular track and is taking the fourth round.

He takes 3 rounds in 40 × 3 = 120 s. Thus, after 120 s his displacement is zero.

Then, the net displacement of the athlete is in 20 s only. In this interval of time, he moves at the opposite end of the initial position. Since displacement is equal to the shortest distance between the initial and final position of the athlete, displacement of the athlete will be equal to the diameter of the circular track.

∴ Displacement of the athlete = 200 m

Distance covered by the athlete in 2 min 20 s is 2200 m and his displacement is
200 m.

 

Answer:

(a) 2 m/s, 2 m/s (b) 1.90 m/s, 0.95 m/s

(a) From end A to end B

Distance covered by Joseph while jogging from A to B = 300 m

Time taken to cover that distance = 2 min 30 seconds = 150 s

Total distance covered = 300 m

Total time taken = 150 s

$\mathrm{Average} \mathrm{speed} = \frac{300}{150 } = 2 \mathrm{m}/\mathrm{s}$

Displacement = shortest distance between A and B = 300 m

Time interval = 150 s

$\mathrm{Average} \mathrm{velocity} = \frac{300}{150 } = 2 \mathrm{m}/\mathrm{s}$

The average speed and average velocity of Joseph from A to B are the same and equal to 2 m/s.

(b) From end A to end C

Total distance covered = Distance from A to B + Distance from B to C

= 300 + 100 = 400 m

Total time taken = Time taken to travel from A to B + Time taken to travel from B to C = 150 + 60 = 210 s

$\mathrm{Average} \mathrm{speed} = \frac{400}{210} = 1.90 \mathrm{m}/\mathrm{s}$

Displacement from A to C = AC = AB − BC = 300 − 100 = 200 m

Time interval = time taken to travel from A to B + time taken to travel from B to C

= 150 + 60 = 210 s

$\mathrm{Average} \mathrm{velocity} = \frac{200}{210} = 0.95 \mathrm{m}/\mathrm{s}$

The average speed of Joseph from A to C is 1.90 m/s and his average velocity is 0.95 m/s.

Answer:

Case I: While driving to school

Average speed of Abdul’s trip = 20 km/h

Total distance = Distance travelled to reach school = d

Let total time taken = t₁

…(i)

Case II: While returning from school

Total distance = Distance travelled while returning from school = d

Now,total time taken = t₂

$$\begin{gathered} 30=\frac{d}{t_2} \\ {t}_2=\frac{d}{30} ..... \left(ii\right) \end{gathered}$$

Where,

Total distance covered in the trip = d + d = 2d

Total time taken, t = Time taken to go to school + Time taken to return to school

= t₁ + t₂

From equations (i) and (ii),

$$\begin{gathered} \mathrm{Average} \mathrm{Speed}=\frac{2\mathrm{d}}{{\displaystyle \frac{\mathrm{d}}{20}+\frac{\mathrm{d}}{30}}}=\frac{2}{{\displaystyle \frac{3+2}{60}}} \\ \mathrm{Average} \mathrm{Speed}=\frac{120}{5}=24 \mathrm{km}/\mathrm{h} \end{gathered}$$

Hence, the average speed for Abdul’s trip is 24 km/h.

Answer:

Initial velocity, u = 0 (since the motor boat is initially at rest)

Acceleration of the motorboat, a = 3 m/s²

Time taken, t = 8 s

According to the second equation of motion:

Distance covered by the motorboat, s

Hence, the boat travels a distance of 96 m.

Answer:

Case A:

Initial speed of the car, u₁ = 52 km/h = 14.4 m/s

Time taken to stop the car, t₁ = 5 s

Final speed of the car becomes zero after 5 s of application of brakes.

Case B:

Initial speed of the car, u₂ = 3 km/h = 0.833 m/s ≅ 0.83 m/s

Time taken to stop the car, t₂ = 10 s

Final speed of the car becomes zero after 10 s of application of brakes.

Plot of the two cars on a speed−time graph is shown in the following figure:

Distance covered by each car is equal to the area under the speed−time graph.

Distance covered in case A,

s₁

Distance covered in case B,

s₂

Area of ΔOPR > Area of ΔOSQ

Thus, the distance covered in case A is greater than the distance covered in case B.

Hence, the car travelling with a speed of 52 km/h travels farther after brakes were applied.

Answer:

(a)

∴Speed = slope of the graph

Since slope of object B is greater than objects A and C, it is travelling the fastest.

(b) All three objects A, B and C never meet at a single point. Thus, they were never at the same point on road.

(c)

On the distance axis:

7 small boxes = 4 km

∴1 small box

Initially, object C is 4 blocks away from the origin.

∴Initial distance of object C from origin=

Distance of object C from origin when B passes A = 8 km

Distance covered by C

Hence, C has travelled a distance of 5.714 km when B passes A.

(d)

Distance covered by B at the time it passes C = 9 boxes

Hence, B has travelled a distance of 5.143 km when B passes A.

Answer:

Distance covered by the ball, s = 20 m

Acceleration, a = 10 m/s²

Initially, velocity, u = 0 (since the ball was initially at rest)

Final velocity of the ball with which it strikes the ground, v

According to the third equation of motion:

v² = u² + 2 as

v² = 0 + 2 (10) (20)

v = 20 m/s

According to the first equation of motion:

v = u + at

Where,

Time, t taken by the ball to strike the ground is,

20 = 0 + 10 (t)

t = 2 s

Hence, the ball strikes the ground after 2 s with a velocity of 20 m/s.

 

Answer:

(a)

The shaded area which is equal to represents the distance travelled by the car in the first 4 s.

(b)

The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.

Answer:

(a) Possible

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) Possible

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

 

Answer:

$$\begin{gathered} \mathrm{Speed} = \frac{\text{Distance}}{\text{Time}} \\ \text{Distance }= 2\pi r\mathit{ }= 2\times 3.14\times 42250 = 265330 \mathrm{km} \\ \text{Time} = 24 \mathrm{h} \\ \text{Speed} = \frac{265330}{24} = 11055.4 \mathrm{km}/\mathrm{h} \end{gathered}$$