Contents

### Access Answers of Science NCERT class 9 Chapter 8: Motion (All intext and exercise questions solved)

**Intext Questions – 1 Page: 100**

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

**Solution**

Yes, an object moving a certain distance can have zero total displacement. Displacement refers to the shortest distance between the initial and the final positions of the object. Even if an object moves through a considerable distance, if it eventually comes back to its initial position, the corresponding displacement of the object would be zero.

**2. A farmer moves along the boundary of a square field of side 10m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?**

**Solution**

Given that the farmer covers the entire boundary of the square field in 40 seconds, the total distance traveled by the farmer in 40 seconds is 4*(10) = 40 meters.

Therefore, the average distance covered by the farmer in one second is: 40m/40 = 1m

Two minutes and 20 seconds can be written as 140 seconds. The total distance traveled by the farmer in this timeframe is: 1 m * 140 = 140m

Since the farmer is moving along the boundary of the square field, the total number of laps completed by the farmer will be: 140m/40 = 3.5 laps

Now, the total displacement of the farmer depends on the initial position. If the initial position of the farmer is at one corner of the field, the terminal position would be at the opposite corner (since the field is square).

In this case, the total displacement of the farmer will be equal to the length of the diagonal line across the opposite corners of the square.

Applying the Pythagoras theorem, the length of the diagonal can be obtained as follows: **√(10 ^{2}+10^{2})**=

**√200**=

**14.14m**.

This is the **maximum** possible displacement of the farmer.

If the initial position of the farmer is at the mid-point between two adjacent corners of the square, the net displacement of the farmer would be equal to the side of the square, which is 10m. This is the **minimum **displacement.

If the farmer starts at a random point around the perimeter of the square, his net displacement after traveling 140m will lie between 10m and 14.14m.

**3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.**

**Solution**

Neither of the statements are true. Statement (a) is false because the displacement of an object which travels a certain distance and comes back to its initial position is zero. Statement (b) is false because the displacement of an object can be equal to, but never greater than the distance traveled.

**Intext Questions – 2 Page: 102**

**1. Distinguish between speed and velocity.**

**Solution**

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

**Solution**

Since average speed is the total distance traveled in a time frame and velocity is the total displacement in the time frame, the magnitude of average velocity and average speed will be the same when the total distance traveled is equal to the displacement.

**3. What does the odometer of an automobile measure?**

**Solution**

The odometer measures the total distance traveled by the automobile.

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

**Solution**

The path of an object in uniform motion is a straight line.

**5. 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 the speed of light, that is, 3 × 10 ^{8} m/s.**

**Solution**

Given that the signal travels in a straight line, the distance between the spaceship and the ground station is equal to the total distance traveled by the signal.

5 minutes = 5*60 seconds = 300 seconds.

Speed of the signal = 3 × 10^{8} m/s.

Therefore, total distance = (3 × 10^{8} m/s) * 300s

= 9*10^{10} meters.

**Intext Questions – 3 Page: 103**

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

**Solution**

**Uniform Acceleration: **In this type of acceleration, the body moves along a straight line and its velocity increases/decreases at a uniform rate (it changes at a constant rate in any constant time interval).

**Non-Uniform Acceleration:** In this type of acceleration, the body moves along a straight line and its velocity increases/decreases at a rate that is not uniform (it changes at a different rate for a given constant time interval).

**2. A bus decreases its speed from 80 km h–1 to 60 km h–1 in 5 s. Find the acceleration of the bus.**

**Solution**

Given, the initial velocity (u) = 80km/hour = 80000m/3600s= 22.22 m.s^{-1}

The final velocity (v) = 60km/hour = 60000m/3600s= 16.66 m.s^{-1}

Time frame, t = 5 seconds.

Therefore, acceleration (a) =(v-u)/t = (16.66 m.s^{-1} – 22.22 m.s^{-1})/5s

= -1.112 m.s^{-2}

Therefore, the total acceleration of the bus is -1.112m.s^{-2}. It can be noted that the negative sign indicates that the velocity of the bus is decreasing.

**3. 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.**

**Solution**

Given, the initial velocity (u) of the train = 0m.s^{-1} (at rest)

Terminal velocity (v) of the train = 40km/hour = 11.11 m.s^{-1}

Time interval, t = 10 minutes = 600 s.

The acceleration of the train is given by =(v-u)/t = (11.11 m.s^{-1} – 0 m.s^{-1})/600s = **0.0185 m.s ^{-2}**

**Intext Questions – 4 Page: 107**

**1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?**

**Solution**

For uniform motion, the distance-time graph is a straight line. On the other hand, the distance-time graph of an object in non-uniform motion is a curve.

The first graph describes uniform motion and the second one describes non-uniform motion.

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

**Solution**

This distance-time graph can be plotted as follows.

Since there is no change in the distance traveled by the object (or the Y-Axis value) at any point in the X-Axis (time), the object is at rest.

**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?**

**Solution**

This speed-time graph can be plotted as follows.

Since there is no change in the velocity of the object (Y-Axis value) at any point of time (X-axis value), the object is said to be in uniform motion.

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

**Solution**

Considering an object in uniform motion, its velocity-time graph can be represented as follows.

Now, the area below the velocity-time graph is the area of the rectangle OABC, which is given by OA*OC. But OA is the velocity of the object and OC represents time. Therefore, the shaded area can be represented as:

Area under the velocity-time graph = velocity*time.

Substituting the value of velocity as displacement/time in the previous equation, it is found that the area under the velocity-time graph represents the total displacement of the object.

**Intext Questions – 5 Page: 109,110**

**1. A bus starting from rest moves with a uniform acceleration of 0.1 m s ^{-2} for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.**

**Solution**

(a) Given, the bust starts from rest. Therefore, initial velocity (u) = 0 m/s

Acceleration (a) = 0.1 m.s^{-2}

Time = 2 minutes = 120 s

Acceleration is given by the equation a=(v-u)/t

Therefore, terminal velocity (v) = (at)+u

= (0.1 m.s^{-2} * 120s) + 0 m.s^{-1}

= 12m.s^{-1} + 0 m.s^{-1}

Therefore, terminal velocity (v) = 12m/s

(b) As per the third motion equation,

Since a = 0.1 m.s^{-2}, v = 12 m.s^{-1}, u = 0 m.s^{-1}, and t = 120s, the following value for s (distance) can be obtained.

Distance, s =(v^{2} – u^{2})/2a

=(12^{2} – 0^{2})/2(0.1)

Therefore, s = 720m.

The speed acquired is 12m.s^{-1} and the total distance traveled is 720m.

**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 m s^{-2}. Find how far the train will go before it is brought to rest.**

**Solution**

Given, initial velocity (u) = 90 km/hour = 25 m.s^{-1}

Terminal velocity (v) = 0 m.s^{-1}

Acceleration (a) = -0.5 m.s^{-2}

As per the third motion equation, v^{2}-u^{2}=2as

Therefore, distance traveled by the train (s) =(v^{2}-u^{2})/2a

s = (0^{2}-25^{2})/2(-0.5) meters = 625 meters

The train must travel 625 meters at an acceleration of -0.5 ms^{-2} before it reaches the rest position.

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

**Solution**

Given, initial velocity (u) = 0 (the trolley begins from the rest position)

Acceleration (a) = 0.02 ms^{-2}

Time (t) = 3s

As per the first motion equation, v=u+at

Therefore, terminal velocity of the trolley (v) = 0 + (0.02 ms^{-2})(3s)= 0.06 ms^{-2}

Therefore, the velocity of the trolley after 3 seconds will be 6 cm.s^{-2}

**4. A racing car has a uniform acceleration of 4 m s ^{-2}. What distance will it cover in 10 s after start?**

**Solution**

Given, the car is initially at rest; initial velocity (u) = 0 ms^{-1}

Acceleration (a) = 4 ms^{-2}

Time period (t) = 10 s

As per the second motion equation, s = ut+1/2 at^{2}

Therefore, the total distance covered by the car (s) = 0 * 10m + 1/2 (4ms^{-2})(10s)^{2}

= 200 meters

Therefore, the car will cover a distance of 200 meters after 10 seconds.

**5. 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?**

**Solution**

Given, initial velocity (u) = 5 m/s

Terminal velocity (v) = 0 m/s (since the stone will reach a position of rest at the point of maximum height)

Acceleration = 10 ms^{-2} in the direction opposite to the trajectory of the stone = -10 ms^{-2}

As per the third motion equation, v^{2} – u^{2} = 2as

Therefore, the distance traveled by the stone (s) = (0^{2} – 5^{2})/ 2(10)

Distance (s) = 1.25 meters

As per the first motion equation, v = u + at

Therefore, time taken by the stone to reach a position of rest (maximum height) = (v – u) /a

=(0-5)/-10 s

Time taken = 0.5 seconds

Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.