---
title: "NCERT Solutions Class 9 Science Ch 4 Describing Motion Around Us"
url: https://www.swavid.com/science/class/9/chapter/describing-motion-around-us/ncert-solutions
dateModified: 2026-10-07T16:51:05+00:00
---

# NCERT Solutions Class 9 Science Ch 4 Describing Motion Around Us

This chapter's questions cover foundational concepts of motion including distance, displacement, speed, velocity, acceleration, graphical representation, kinematic equations, and circular motion.

Free PDF (19 pages): https://www.swavid.com/api/seo/pdf/ncert/science/class-9/swavid-ncert-solutions-class-9-science-chapter-4-describing-motion-around-us-b71e29c9a3.pdf

## Think It Over

### Question 1

*2 marks · Very short answer*

How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes?

**Solution**

1. We should maintain a safe distance from the truck ahead to allow enough stopping distance in case it suddenly applies the brakes.
2. This distance depends on the vehicle's speed, road conditions, braking capacity, and the driver's reaction time.

**Answer:** We must maintain a safe distance based on our speed and stopping capability to avoid a collision.

> Common mistake: Stating an arbitrary fixed distance without considering speed and reaction time.

### Question 2

*2 marks · Very short answer*

Does this distance depend upon the speed with which we are moving?

**Solution**

1. Yes, the safe distance depends directly upon the speed with which we are moving.
2. A higher initial speed increases the stopping distance according to the kinematic equation $v^2 = u^2 + 2as$, requiring a larger gap.

**Answer:** Yes, the distance depends upon our speed because higher speeds require a larger stopping distance.

> Common mistake: Assuming stopping distance is independent of speed.

## Activity 4.1: Let us analyse

### Question 1

*2 marks · Very short answer*

As shown in Fig. 4.5, a ball is thrown vertically upwards from O. It moves up straight till B and then falls back to O. Can this be considered a motion in a straight line?

**Solution**

1. Yes, the motion of the ball thrown vertically upwards and falling back along the same path can be considered motion in a straight line.

**Answer:** Yes, it can be considered a motion in a straight line.

> Common mistake: Thinking that motion becomes two-dimensional because the ball goes up and comes down.

### Question 2

*3 marks · Short answer*

For this motion, fill up the values in Table 4.1.

**Part (i)**

1. At position B: Total distance travelled is $80\text{ cm}$ and displacement is $80\text{ cm}$ in upward direction.

Answer (i): Distance = $80\text{ cm}$, Displacement = $80\text{ cm}$ in upward direction

**Part (ii)**

1. At position C: Total distance travelled is $80\text{ cm} + 20\text{ cm} = 100\text{ cm}$ and displacement is $60\text{ cm}$ in upward direction.

Answer (ii): Distance = $100\text{ cm}$, Displacement = $60\text{ cm}$ in upward direction

**Part (iii)**

1. At final position O (returning back): Total distance travelled is $140\text{ cm}$ and displacement is $0\text{ cm}$.

Answer (iii): Distance = $140\text{ cm}$, Displacement = $0\text{ cm}$

**Answer:** Table 4.1 completed with distance and displacement values for positions O, A, B, C, and O.

> Common mistake: Confusing total distance travelled with displacement when the object turns back.

### Question 3

*1 mark · MCQ*

Analyse the data filled in Table 4.1 and choose which of the following is true for displacement:

- It is never zero.
- Its magnitude can be greater than the total distance travelled.
- Its magnitude is less than or equal to the total distance travelled.
- Its magnitude is less than the total distance travelled in all cases.

**Solution**

1. From Table 4.1, when the ball returns to the starting point O, the displacement is zero, so it can be zero.
2. The magnitude of displacement is the straight-line distance between initial and final positions, which is always less than or equal to the total distance travelled.

**Answer:** (iii) Its magnitude is less than or equal to the total distance travelled.

> Common mistake: Selecting that displacement is never zero, forgetting that a round trip results in zero displacement.

## Pause and Ponder

### Question 1

*3 marks · Short answer*

In the example of an athlete running back and forth on a straight track (Fig. 4.4), when will the displacement of the athlete be zero? What will be the total distance travelled in that case?

**Part (i)**

1. The displacement of an object is defined as the net change in its position between two given instants of time.
2. When the athlete starts from point O and returns back to point O after running back and forth, her initial and final positions are the same.
3. Therefore, the net change in position is zero, making the displacement zero.

Answer (i): The displacement is zero when the athlete returns to the starting position O.

**Part (ii)**

1. The total distance travelled is the actual path length covered by the athlete during the entire motion.
2. Since the athlete returns to O after running back and forth along the track, the total distance travelled depends on the entire path covered, which is non-zero and greater than zero.

Answer (ii): The total distance travelled is the total length of the path covered by the athlete during the entire motion.

**Answer:** The displacement of the athlete will be zero when the athlete returns to the starting point O.

> Common mistake: Confusing displacement with distance travelled when an object returns to its starting point.

### Question 2

*1 mark · MCQ*

Fuel used up in a vehicle depends on which of the following? Justify your answer.

- Total distance travelled
- Displacement

**Solution**

1. Fuel consumption in a vehicle depends on the total path length covered by the vehicle during its journey.
2. Since total distance travelled accounts for the entire path length whereas displacement only considers the net change in position (which can be zero for a round trip), fuel used depends on total distance travelled.
3. Therefore, option (i) is correct.

**Answer:** (i) Total distance travelled

> Common mistake: Thinking fuel depends on displacement, ignoring that a vehicle consumes fuel even during a round trip where displacement is zero.

### Question 3

*3 marks · Short answer*

A ball rolls down an inclined track as shown in Fig. 4.6. Is its motion, a straight line motion? Assuming the starting point of the ball (O) to be the origin, can its motion from O to D be depicted using a horizontal line as shown in Fig. 4.3? Are the values of total distance travelled and magnitude of displacement from O equal or different at positions A, B, C and D?

**Part (i)**

1. The ball rolls down the track shown in Fig. 4.6 along a single straight path.
2. Therefore, its motion is a straight line motion.

Answer (i): Yes, its motion is a straight line motion.

**Part (ii)**

1. Assuming the starting point O to be the origin, the path of motion is along a straight line.
2. Thus, its motion from O to D can be depicted using a horizontal line as shown in Fig. 4.3.

Answer (ii): Yes, its motion can be depicted using a horizontal line.

**Part (iii)**

1. The ball moves from O to D in one direction without turning back along the straight track.
2. For motion in a straight line in one direction, the total distance travelled and the magnitude of displacement are equal at positions A, B, C, and D.

Answer (iii): The values of total distance travelled and magnitude of displacement are equal at positions A, B, C, and D.

**Answer:** The motion is along a straight inclined track, its motion can be depicted on a straight line, and the values of total distance and magnitude of displacement are equal.

> Common mistake: Assuming distance and displacement magnitude differ when the motion is along an incline in one direction.

## Pause and Ponder

### Question 4

*3 marks · Numerical*

During a family road trip, you drive 200 km north in three hours. Afterwards, you drive 200 km south in two hours. Find the average speed and average velocity for your entire trip.

**Solution**

1. Given: Total distance travelled = $200\text{ km} + 200\text{ km} = 400\text{ km}$, total time interval = $3\text{ h} + 2\text{ h} = 5\text{ h}$, and net displacement = $0\text{ km}$ since the final position is the same as the starting point.
2. Formula: $\text{average speed} = \frac{\text{total distance travelled}}{\text{time interval}}$ and $\text{average velocity} = \frac{\text{displacement}}{\text{time interval}}$.
3. Substitution: $\text{average speed} = \frac{400\text{ km}}{5\text{ h}}$ and $\text{average velocity} = \frac{0\text{ km}}{5\text{ h}}$.
4. Result: Average speed = $80\text{ km h}^{-1}$ and average velocity = $0\text{ km h}^{-1}$.

**Answer:** Average speed = $80\text{ km h}^{-1}$, average velocity = $0\text{ km h}^{-1}$

> Common mistake: Students often divide the sum of speeds or use displacement instead of total distance while calculating average speed.

### Question 5

*3 marks · Short answer*

Under what condition(s) is the (i) magnitude of average velocity of an object equal to its average speed? (ii) magnitude of average velocity of an object zero while its average speed is not zero?

**Part (i)**

1. The magnitude of average velocity is equal to average speed when an object moves along a straight line without turning back, so that distance travelled and displacement are equal.

Answer (i): When the object moves in a straight line in one direction.

**Part (ii)**

1. The magnitude of average velocity is zero when the net displacement is zero (starting and ending points are the same), while the total distance travelled is non-zero.

Answer (ii): When the object returns to its starting point after travelling some distance.

**Answer:** The magnitude of average velocity equals average speed when the object moves in a straight line in one direction. The magnitude of average velocity is zero while average speed is not zero when the object returns to its starting point after a journey.

> Common mistake: Thinking that average velocity can be non-zero when starting and ending points coincide.

## Activity 4.2: Let us calculate

### Question 1

*Activity*

The magnitude of average acceleration of cars is generally specified as the time taken by the car to go from $0 \text{ km h}^{-1}$ to $100 \text{ km h}^{-1}$. Look it up on the internet and find this time for various cars, and record those in Table 4.2.

**Solution**

1. This is a classroom activity where students research the 0 to 100 km h-1 acceleration time for various cars from reliable sources.
2. Record the car type and the corresponding time interval in Table 4.2.

**Answer:** Data recorded in Table 4.2 based on internet research.

### Question 2

*Activity*

Calculate the magnitude of average acceleration for each car.

**Solution**

1. Convert the final velocity from km h-1 to m s-1 by multiplying by $\frac{1000\text{ m}}{3600\text{ s}}$ so that $100\text{ km h}^{-1} = 27.8\text{ m s}^{-1}$.
2. Use the formula $a = \frac{v - u}{t}$, where $u = 0\text{ m s}^{-1}$, $v = 27.8\text{ m s}^{-1}$, and $t$ is the time interval recorded in Table 4.2.
3. Calculate the magnitude of average acceleration for each car and record the values in Table 4.2.

**Answer:** Magnitude of average acceleration calculated and filled in Table 4.2.

## Activity 4.3: Let us plot a graph

### Question 1

*Activity*

Take a sheet of graph paper. This paper is pre-divided into small squares (Fig. 4.11a), making it easier to plot data accurately.

**Solution**

1. Take a pre-divided square grid graph paper as shown in Fig. 4.11a to plot motion data accurately.

**Answer:** A graph paper pre-divided into small squares is taken for accurate plotting.

### Question 2

*Activity*

On the graph paper, draw two lines perpendicular to each other as shown in Fig. 4.11a. Their point of intersection is known as origin O. Mark the horizontal line as OX. It is known as the X-axis. Similarly, mark the vertical line as OY. It is called the Y-axis.

**Solution**

1. Draw two mutually perpendicular lines on the graph paper whose intersection point is the origin O.
2. Mark the horizontal line as OX (X-axis) and the vertical line as OY (Y-axis).

**Answer:** Perpendicular axes X-axis and Y-axis intersecting at origin O are drawn.

### Question 3

*Activity*

Refer to Table 4.3. We need to decide which quantity (time or position) to be shown along each axis. For the data we have (Table 4.3), we will show time along the X-axis and position along the Y-axis.

**Solution**

1. Refer to Table 4.3 to identify time and position values.
2. Assign time along the horizontal X-axis and position along the vertical Y-axis.

**Answer:** Time is chosen along the X-axis and position along the Y-axis.

### Question 4

*Activity*

Determine a suitable scale for each quantity to represent it on the graph paper. We need to choose scales that allow us to represent the data effectively and conveniently while utilising the available space.

**Solution**

1. Choose a suitable scale to represent the given data effectively utilising the available space.
2. Use X-axis: 5 divisions = $1\text{ s}$ and Y-axis: 5 divisions = $20\text{ m}$.

**Answer:** Scales chosen are 5 divisions = $1\text{ s}$ for X-axis and 5 divisions = $20\text{ m}$ for Y-axis.

### Question 5

*Activity*

Use the chosen scale to mark values for time ($1\text{ s}, 2\text{ s}, \dots$) along the X-axis from the origin. Similarly, mark values for position ($20\text{ m}, 40\text{ m}, \dots$) along the Y-axis (Fig. 4.11b).

**Solution**

1. Mark time values ($1\text{ s}, 2\text{ s}, \dots$) along the X-axis from the origin using the chosen scale.
2. Mark position values ($20\text{ m}, 40\text{ m}, \dots$) along the Y-axis as shown in Fig. 4.11b.

**Answer:** Values of time and position are marked along the X-axis and Y-axis respectively.

### Question 6

*Activity*

Begin plotting points on the graph paper to represent each set of time and position values from Table 4.3.

**Solution**

1. Plot each set of time and position values from Table 4.3 by locating the intersection of lines parallel to the axes.
2. Mark points for $(0\text{ s}, 0\text{ m})$, $(1\text{ s}, 20\text{ m})$, $(2\text{ s}, 40\text{ m})$, and subsequent readings on the graph paper.

**Answer:** Points corresponding to each set of time and position values from Table 4.3 are plotted on the graph paper.

### Question 7

*Activity*

Once all points are plotted, connect them to create the position-time graph for the vehicle's motion (Fig. 4.11c).

**Solution**

1. Refer to the plotted data points representing the position of the vehicle at different instants of time as given in Table 4.3.
2. Join all the plotted points smoothly using a ruler to create the position-time graph as shown in Fig. 4.11(c).
3. Observe that the resulting graph is a straight line, which indicates uniform motion in a straight line with constant speed.

**Answer:** The points are connected to obtain a straight-line position-time graph (Fig. 4.11c).

> Common mistake: Freehand sketching instead of using a ruler to join the points representing a straight-line graph.

## Activity 4.4: Let us calculate

### Question 1

*Activity*

In the position-time graph we plotted (Fig. 4.11c), consider a part (say, AB) of the graph as shown in Fig. 4.14. From A, draw a line parallel to X-axis and another line parallel to Y-axis. Repeat the same from B.

**Solution**

1. From point A on the position-time graph, draw a line parallel to the X-axis and another line parallel to the Y-axis.
2. Repeat the exact same procedure from point B to complete the geometric construction for finding velocity.

**Answer:** Lines parallel to the X-axis and Y-axis are drawn from points A and B on the position-time graph.

### Question 2

*Activity*

Extend the horizontal line from A and a triangle ABC is formed. What do the sides BC and CA of the triangle represent?

**Solution**

1. Extending the horizontal line from A forms a right-angled triangle ABC.
2. Side BC represents the change in position ($s_2 - s_1$), and side CA represents the change in time ($t_2 - t_1$).

**Answer:** Side BC represents change in position ($s_2 - s_1$) and side CA represents change in time ($t_2 - t_1$).

### Question 3

*Activity*

As per Eq. (4.2a), by dividing the change in position (BC) by the change in time (CA), you get the average velocity.

**Solution**

1. Divide the change in position (BC) by the change in time (CA) as per Eq. (4.2a).
2. This ratio $\frac{\text{BC}}{\text{CA}}$ gives the average velocity, which is also the slope of the straight line AB.

**Answer:** $v = \frac{s_2 - s_1}{t_2 - t_1} = \frac{\text{BC}}{\text{CA}}$

### Question 4

*Activity*

By extracting values of time $t_1$ and $t_2$, and distances $s_1$ and $s_2$ from the graph, the magnitude of average velocity can be calculated.

**Solution**

1. Extract the values from Fig. 4.14: $t_1 = 2\text{ s}$, $t_2 = 4\text{ s}$, $s_1 = 40\text{ m}$, and $s_2 = 80\text{ m}$.
2. Substitute these values into the formula to calculate the magnitude of average velocity: $v = \frac{80\text{ m} - 40\text{ m}}{4\text{ s} - 2\text{ s}} = \frac{40\text{ m}}{2\text{ s}} = 20\text{ m s}^{-1}$.

**Answer:** $20\text{ m s}^{-1}$

## Activity 4.5: Let us investigate

### Question 1

*Activity*

Take a ring, such as an adhesive tape ring and one marble.

**Solution**

1. Take an adhesive tape ring and a marble.

**Answer:** Apparatus is ready for the activity.

### Question 2

*Activity*

Place the ring flat on a smooth surface and throw the marble inside the ring in a way that it rotates along the inner boundary of the ring (Fig. 4.24).

**Solution**

1. Place the ring flat on a smooth surface.
2. Throw the marble inside the ring so that it rotates along the inner boundary.

**Answer:** The marble rotates along the inner boundary of the ring in circular motion.

### Question 3

*Activity*

Predict what will happen if you lift the ring while the marble is moving.

**Solution**

1. Predict that the marble will not continue in a circle but will move away along a straight line.

**Answer:** The marble will move in a straight line tangent to the circular path.

### Question 4

*Activity*

Now, after one or two complete revolutions of the marble, pick up the ring without disturbing the motion of the marble. What do you observe? Does the marble continue moving in a circular motion? Or does it move in some other manner?

**Solution**

1. Allow the marble to complete one or two revolutions inside the ring.
2. Pick up the ring without disturbing the marble and observe its path.

**Answer:** The marble does not continue in a circular motion; it moves out in a straight line.

### Question 5

*Activity*

Repeat the activity multiple times to confirm the result.

**Solution**

1. Repeat the activity multiple times to confirm the observation.

**Answer:** Repeated trials consistently show that the released marble moves along a straight line.

## Revise, Reflect, Refine

### Question 1

*3 marks · Numerical*

My father went to a shop from home which is located at a distance of $250\text{ m}$ on a straight road. On reaching there, he discovered that he forgot to carry a cloth bag. He came home to take it, went to the shop again, bought provisions and came back home. How much was the total distance travelled by him? What was his displacement from home?

**Solution**

1. Given: Distance from home to the shop = $250\text{ m}$. Total journey consists of: home to shop ($250\text{ m}$), shop to home ($250\text{ m}$), home to shop again ($250\text{ m}$), and shop back home ($250\text{ m}$).
2. Total distance travelled = $250\text{ m} + 250\text{ m} + 250\text{ m} + 250\text{ m} = 1000\text{ m}$
3. Displacement is the net change in position between the initial and final positions. Since the starting point (home) and the final stopping point (home) are the same, the net change in position is zero.
4. Result: Total distance = $1000\text{ m}$, Displacement = $0\text{ m}$.

**Answer:** Total distance travelled = $1000\text{ m}$, Displacement = $0\text{ m}$

> Common mistake: Confusing total path length (distance) with displacement, or assuming displacement is non-zero when returning to the starting point.

### Question 2

*3 marks · Case-based*

A student runs from the ground floor to the fourth floor of a school building to collect a book and then comes down to their classroom on the second floor. If the height of each floor is $3\text{ m}$, find: (i) the total vertical distance travelled, and (ii) their displacement from the starting point.

**Part (i)**

1. The student runs from the ground floor to the fourth floor, covering $4 - 0 = 4$ floors upwards, and then comes down from the fourth floor to the second floor, covering $4 - 2 = 2$ floors downwards.
2. Total floors covered = $4 + 2 = 6$ floors.
3. Given the height of each floor is $3\text{ m}$, total vertical distance travelled = $6 \times 3\text{ m} = 18\text{ m}$.

Answer (i): $18\text{ m}$

**Part (ii)**

1. Displacement is the net change in position between the starting point (ground floor, $0\text{ m}$) and the final position (second floor, $2\text{ nd}$ floor).
2. Net floors from the ground = $2$ floors.
3. Displacement = $2 \times 3\text{ m} = 6\text{ m}$ in the upward direction.

Answer (ii): $6\text{ m}$ in the upward direction

**Answer:** (i) Total vertical distance travelled = $18\text{ m}$, (ii) Displacement = $6\text{ m}$ upwards.

> Common mistake: Confusing total path length with net displacement by subtracting downward distance from distance travelled.

### Question 3

*3 marks · Short answer*

A girl is riding her scooter and finds that its speedometer reading is constant. Is it possible for her scooter to be accelerating and if so, how?

**Solution**

1. Yes, it is possible for the scooter to be accelerating even if its speedometer reading (speed) is constant.
2. Acceleration depends on a change in velocity, which can be due to a change in the magnitude of velocity, its direction, or both.
3. If the girl is riding along a curved path or taking a turn, the direction of velocity changes continuously at every instant, making the motion accelerated (uniform circular motion).

**Answer:** Yes, the scooter is accelerating if it is taking a turn, because acceleration occurs due to the continuous change in the direction of velocity even when speed is constant.

> Common mistake: Assuming acceleration is zero simply because the speedometer reading does not change.

### Question 4

*3 marks · Numerical*

A car starts from rest and its velocity reaches $24\text{ m s}^{-1}$ in $6\text{ s}$. Find the average acceleration and the distance travelled in these $6\text{ s}$.

**Solution**

1. Given: Initial velocity $u = 0\text{ m s}^{-1}$, final velocity $v = 24\text{ m s}^{-1}$, time interval $t = 6\text{ s}$.
2. Formula: $a = \frac{v - u}{t}$
3. Substitution: $a = \frac{24\text{ m s}^{-1} - 0\text{ m s}^{-1}}{6\text{ s}} = \frac{24}{6} = 4\text{ m s}^{-2}$.
4. Formula for distance: $s = ut + \frac{1}{2}at^2$.
5. Substitution: $s = (0\text{ m s}^{-1} \times 6\text{ s}) + \frac{1}{2} \times 4\text{ m s}^{-2} \times (6\text{ s})^2 = 0 + 2 \times 36 = 72\text{ m}$.
6. Result: Average acceleration = $4\text{ m s}^{-2}$, Distance travelled = $72\text{ m}$.

**Answer:** Average acceleration = $4\text{ m s}^{-2}$, Distance travelled = $72\text{ m}$

> Common mistake: Using incorrect kinematic equations or forgetting to square the time in the distance formula.

### Question 5

*3 marks · Numerical*

A motorbike moving with initial velocity $28\text{ m s}^{-1}$ and constant acceleration stops after travelling $98\text{ m}$. Find the acceleration of the motorbike and the time taken to come to a stop.

**Solution**

1. Given: Initial velocity $u = 28\text{ m s}^{-1}$, final velocity $v = 0\text{ m s}^{-1}$ (stops), distance $s = 98\text{ m}$.
2. Formula: $v^2 = u^2 + 2as$
3. Substitution: $(0)^2 = (28)^2 + 2 \times a \times 98 \implies 0 = 784 + 196a \implies a = \frac{-784}{196} = -4\text{ m s}^{-2}$.
4. Formula for time: $v = u + at \implies 0 = 28 + (-4)t \implies 4t = 28 \implies t = 7\text{ s}$.
5. Result: Acceleration = $-4\text{ m s}^{-2}$, Time taken = $7\text{ s}$.

**Answer:** Acceleration = $-4\text{ m s}^{-2}$, Time taken = $7\text{ s}$

> Common mistake: Omitting the negative sign for acceleration when the vehicle comes to a stop.

### Question 6

*3 marks · Short answer*

Fig. 4.27 shows a position-time graph of two objects A and B that are moving along the parallel tracks in the same direction. Do objects A and B ever have equal velocity? Justify your answer.

**Solution**

1. In a position-time graph, the slope of the graph represents the velocity of the object.
2. Referring to the figure in the textbook (Fig. 4.27), the two lines representing the motion of objects A and B have different slopes at various points, and at no point do the tangents or slopes of the two lines become equal.
3. Therefore, objects A and B never have equal velocity at the same instant of time.

**Answer:** No, objects A and B never have equal velocity because their position-time graph lines have different slopes at all corresponding instants of time.

> Common mistake: Confusing position with velocity, or assuming equal positions mean equal velocities.

### Question 7

*1 mark · MCQ*

A graph in Fig. 4.28 shows the change in position with time for two objects A and B moving in a straight line from 0 to 10 seconds. Choose the correct option(s).

- The average velocity of both over the $10\text{ s}$ time interval is equal since they have the same initial and final positions.
- The average speeds of both over the $10\text{ s}$ time interval are equal since both cover equal distance in equal time.
- The average speed of A over the $10\text{ s}$ time interval is lower than that of B since it covers a shorter distance than B in 10 seconds.
- The average speed of A over the $10\text{ s}$ time interval is greater than that of B since B's speed is lower than A's in some segments.

**Solution**

1. From the position-time graph (Fig. 4.28), object A covers a larger vertical change in position than object B in the same $10\text{ s}$ time interval.
2. Since average speed is total distance divided by time interval, and both move in one direction, object A has a greater average speed than object B.
3. Therefore, option (iii) is the correct statement because object A covers a shorter distance is false; A covers more distance.

**Answer:** (iii) The average speed of A over the $10\text{ s}$ time interval is lower than that of B since it covers a shorter distance than B in 10 seconds.

> Common mistake: Confusing the slopes of position-time graphs and assuming the lower curve represents higher speed.

### Question 8

*3 marks · Numerical*

A truck driver driving at the speed of $54\text{ km h}^{-1}$ notices a road sign with a speed limit of $40\text{ km h}^{-1}$ (Fig. 4.29) for trucks. He slows down to $36\text{ km h}^{-1}$ in $36\text{ s}$. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.

**Solution**

1. Given: initial velocity $u = 54\text{ km h}^{-1} = 54 \times \frac{1000\text{ m}}{3600\text{ s}} = 15\text{ m s}^{-1}$, final velocity $v = 36\text{ km h}^{-1} = 10\text{ m s}^{-1}$, time interval $t = 36\text{ s}$.
2. Formula: acceleration $a = \frac{v - u}{t}$ and distance $s = ut + \frac{1}{2}at^2$.
3. Substitution: $a = \frac{10\text{ m s}^{-1} - 15\text{ m s}^{-1}}{36\text{ s}} = \frac{-5}{36}\text{ m s}^{-2}$.
4. Substitution for distance: $s = (15\text{ m s}^{-1})(36\text{ s}) + \frac{1}{2}\left(-\frac{5}{36}\text{ m s}^{-2}\right)(36\text{ s})^2 = 540 - 90 = 450\text{ m}$.
5. Result: $450\text{ m}$.

**Answer:** $450\text{ m}$

> Common mistake: Forgetting to convert km/h to m/s before performing calculations.

### Question 9

*3 marks · Numerical*

A car starts from rest and accelerates uniformly to $20\text{ m s}^{-1}$ in 5 seconds. It then travels at $20\text{ m s}^{-1}$ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.

**Solution**

1. Stage 1: Car starts from rest ($u = 0$) and accelerates uniformly to $v = 20\text{ m s}^{-1}$ in $t_1 = 5\text{ s}$. Distance $s_1 = \frac{u + v}{2} \times t_1 = \frac{0 + 20}{2} \times 5 = 50\text{ m}$.
2. Stage 2: Car travels at a constant velocity of $20\text{ m s}^{-1}$ for $t_2 = 10\text{ s}$. Distance $s_2 = \text{velocity} \times \text{time} = 20 \times 10 = 200\text{ m}$.
3. Stage 3: Car applies brakes and stops ($v = 0$) from $u = 20\text{ m s}^{-1}$ in $t_3 = 6\text{ s}$. Distance $s_3 = \frac{u + v}{2} \times t_3 = \frac{20 + 0}{2} \times 6 = 60\text{ m}$.
4. Total distance travelled $s = s_1 + s_2 + s_3 = 50\text{ m} + 200\text{ m} + 60\text{ m} = 310\text{ m}$.
5. Result: $310\text{ m}$.

**Answer:** $310\text{ m}$

> Common mistake: Using incorrect kinematic formulas or mixing up time intervals for the three distinct stages of motion.

### Question 10

*3 marks · Numerical*

A bus is travelling at $36\text{ km h}^{-1}$ when the driver sees an obstacle $30\text{ m}$ ahead. The driver takes 0.5 seconds to react before pressing the brake. Once the brake is applied, the velocity of the bus reduces with constant acceleration of $2.5\text{ m s}^{-2}$. Will the bus be able to stop before reaching the obstacle?

**Solution**

1. Given: Initial velocity $u = 36\text{ km h}^{-1} = 10\text{ m s}^{-1}$, reaction time $t_r = 0.5\text{ s}$, deceleration $a = -2.5\text{ m s}^{-2}$, distance to obstacle $d = 30\text{ m}$.
2. Distance travelled during reaction time ($s_1$): $s_1 = u \times t_r = 10\text{ m s}^{-1} \times 0.5\text{ s} = 5\text{ m}$.
3. Distance travelled after brakes are applied ($s_2$): Using $v^2 = u^2 + 2as$, $0 = (10)^2 + 2(-2.5)s_2$, giving $5s_2 = 100$, so $s_2 = 20\text{ m}$.
4. Total distance travelled before stopping $s_{\text{total}} = s_1 + s_2 = 5\text{ m} + 20\text{ m} = 25\text{ m}$.
5. Result: Since $25\text{ m} < 30\text{ m}$, the bus will be able to stop before reaching the obstacle.

**Answer:** Yes, the bus will be able to stop before reaching the obstacle as the total stopping distance is $25\text{ m}$.

> Common mistake: Ignoring the distance travelled during the driver's reaction time.

### Question 11

*3 marks · Short answer*

A student said, "The Earth moves around the Sun". In this context, discuss whether an object kept on the Earth can be considered to be at rest.

**Solution**

1. Whether an object is at rest or in motion depends entirely on the chosen reference point.
2. An object kept on the Earth is at rest with respect to another object on the Earth because its position does not change with time relative to the Earth.
3. However, with respect to the Sun or space, the Earth and the object on it are in motion; hence, rest is not absolute.

**Answer:** An object kept on the Earth can be considered at rest with respect to the Earth, but it is in motion with respect to the Sun or other objects outside the Earth.

> Common mistake: Stating that the object is purely at rest without specifying the reference point.

### Question 12

*3 marks · Numerical*

The velocity-time graph from $0\text{ s}$ to $120\text{ s}$ for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist (i) while cyclist is moving with constant velocity. (ii) when the velocity of cyclist is decreasing. Also, calculate the displacement and average acceleration in the $120\text{ s}$ time interval.

**Solution**

1. Given: Velocity-time graph of a cyclist from $0\text{ s}$ to $120\text{ s}$ (Fig. 4.30).
2. Part (i) & (ii): The displacement is represented by the area under the velocity-time graph; the constant velocity region is from $20\text{ s}$ to $60\text{ s}$ and decreasing velocity is from $60\text{ s}$ to $120\text{ s}$.
3. Total displacement = Area of the trapezium under the graph = $\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} = \frac{1}{2} \times (120 + 40) \times 4 = 320\text{ m}$.
4. Average acceleration over the $120\text{ s}$ interval = $\frac{v - u}{t_2 - t_1} = \frac{0 - 0}{120 - 0} = 0\text{ m s}^{-2}$ (since initial and final velocities are zero).
5. Result: Displacement is $320\text{ m}$ and average acceleration over the entire interval is $0\text{ m s}^{-2}$.

**Answer:** Displacement is $320\text{ m}$ and average acceleration is $0\text{ m s}^{-2}$.

> Common mistake: Calculating acceleration over the whole interval using intermediate peaks instead of initial and final points.

### Question 13

*3 marks · Numerical*

A girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the distance she ran based on the graph.

**Solution**

1. Given: The velocity-time graph (Fig. 4.31) with time on the X-axis in hours and velocity on the Y-axis in $\text{km h}^{-1}$.
2. Formula: Distance travelled is equal to the total area enclosed between the velocity-time graph and the time axis.
3. Substitution: From Fig. 4.31, the graph extends up to $t = 7\text{ h}$, with varying speeds around $7.5\text{ km h}^{-1}$. Counting the approximate area or trapezoidal strips under the curve, the total area is roughly $50\text{ km}$.
4. Result: $50\text{ km}$

**Answer:** $50\text{ km}$

> Common mistake: Forgetting to convert units or misreading the grid scale on the velocity-time graph.

### Question 14

*3 marks · Numerical*

On entering a state highway, a car continues to move with a constant velocity of $6\text{ m s}^{-1}$ for 2 minutes and then accelerates with a constant acceleration $1\text{ m s}^{-2}$ for 6 seconds. Find the displacement of the car on the state highway in the $2\text{ min } 6\text{ s}$ time interval by drawing a velocity-time graph for its motion.

**Solution**

1. Given: Initial constant velocity $u = 6\text{ m s}^{-1}$ for $t_1 = 2\text{ minutes} = 120\text{ s}$, then constant acceleration $a = 1\text{ m s}^{-2}$ for $t_2 = 6\text{ s}$.
2. Formula: Displacement is given by the total area under the velocity-time graph, consisting of a rectangle and a trapezium (or triangle).
3. Substitution: Area of rectangle = $6\text{ m s}^{-1} \times 120\text{ s} = 720\text{ m}$. Area during acceleration: initial velocity = $6\text{ m s}^{-1}$, final velocity = $6 + (1 \times 6) = 12\text{ m s}^{-1}$, time = $6\text{ s}$, so area = $\frac{1}{2} \times (6 + 12) \times 6 = 54\text{ m}$.
4. Result: $774\text{ m}$

**Answer:** $774\text{ m}$

> Common mistake: Not converting the time in minutes to seconds before calculating displacement.

### Question 15

*3 marks · Numerical*

Two cars A and B start moving with a constant acceleration from rest, in a straight line. Car A attains a velocity of $5\text{ m s}^{-1}$ in $5\text{ s}$. Car B attains a velocity of $3\text{ m s}^{-1}$ in $10\text{ s}$. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement in the two time intervals mentioned (Hint: Calculate the acceleration in both cases. Then calculate their velocities at five instants of time to plot the graph).

**Solution**

1. Given: Car A attains $5\text{ m s}^{-1}$ in $5\text{ s}$ from rest ($a_A = 1\text{ m s}^{-2}$); Car B attains $3\text{ m s}^{-1}$ in $10\text{ s}$ from rest ($a_B = 0.3\text{ m s}^{-2}$).
2. Formula: Displacement is the area under the velocity-time graph for each car.
3. Substitution: For Car A, displacement in $5\text{ s}$ = $\frac{1}{2} \times 5\text{ s} \times 5\text{ m s}^{-1} = 12.5\text{ m}$. For Car B, displacement in $10\text{ s}$ = $\frac{1}{2} \times 10\text{ s} \times 3\text{ m s}^{-1} = 15\text{ m}$.
4. Result: Displacement of Car A is $12.5\text{ m}$ and of Car B is $15\text{ m}$

**Answer:** Car A: $12.5\text{ m}$, Car B: $15\text{ m}$

> Common mistake: Confusing the areas of triangles with rectangles when calculating displacement from v-t graphs.

### Question 16

*5 marks · Case-based*

Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minute's hand of the wall clock. During the given time interval, what is its: (i) distance travelled, (ii) displacement, (iii) speed, and (iv) velocity. The length of the minute's hand is $7\text{ cm}$ (Fig. 4.32).

**Part (i)**

1. Time interval from 6 PM to 7:30 PM is $1.5\text{ hours}$ or $90\text{ minutes}$, which means the minute's hand completes $1.5$ revolutions.
2. Radius of the circular path $r = 7\text{ cm}$.
3. Distance travelled in one revolution = $2\pi r = 2 \times \frac{22}{7} \times 7\text{ cm} = 44\text{ cm}$.
4. Total distance travelled in $1.5$ revolutions = $1.5 \times 44\text{ cm} = 66\text{ cm}$.

Answer (i): $66\text{ cm}$

**Part (ii)**

1. After completing $1.5$ revolutions, the tip of the minute's hand is diametrically opposite to its starting position.
2. The net change in position (displacement) from the starting point is equal to the diameter of the circle.
3. Displacement = $2 \times r = 2 \times 7\text{ cm} = 14\text{ cm}$.

Answer (ii): $14\text{ cm}$

**Part (iii)**

1. Time interval $t = 90\text{ minutes} = 90 \times 60\text{ s} = 5400\text{ s}$.
2. Average speed = $\frac{\text{Total distance travelled}}{\text{Time interval}}$.
3. Speed = $\frac{66\text{ cm}}{5400\text{ s}} = 0.0122\text{ cm s}^{-1}$.

Answer (iii): $0.0122\text{ cm s}^{-1}$

**Part (iv)**

1. Average velocity = $\frac{\text{Displacement}}{\text{Time interval}}$.
2. Velocity = $\frac{14\text{ cm}}{5400\text{ s}} = 0.00259\text{ cm s}^{-1}$ in the direction from the initial position to the final position.

Answer (iv): $0.00259\text{ cm s}^{-1}$

**Answer:** (i) Distance = $44\text{ cm}$, (ii) Displacement = $0\text{ cm}$, (iii) Speed = $0.0815\text{ cm s}^{-1}$, (iv) Velocity = $0\text{ cm s}^{-1}$.

> Common mistake: Assuming displacement is zero for $1.5$ revolutions, forgetting that $1.5$ complete turns end up at the opposite point of the diameter.

## Frequently asked questions

### How many total questions are there in the NCERT Solutions for Class 9 Science Chapter 4 Describing Motion Around Us?

This chapter features a comprehensive set of questions divided across various learning sections based on the new NCERT book for the 2026-27 session. You can find SwaVid's free PDF and step-by-step solutions for all these questions on this page only.

### Which major topics and activities do these Class 9 Science Chapter 4 solutions cover?

The solutions cover important concepts such as distance and displacement, average speed and velocity, average acceleration, position-time graphs, and circular motion. They thoroughly address the exercises from Activity 4.1 through Activity 4.5, Think It Over, Pause and Ponder, and Revise, Reflect, Refine.

### Which question types are considered the most challenging in this chapter and how should students approach them?

Numerical problems and graphical analysis questions, such as calculating average velocity from position-time graphs or finding the area under velocity-time graphs, are often the hardest. To score full marks, students should carefully label their graph axes, write proper SI units, and clearly state the formula before substituting values.

### How should students write their answers to secure full marks in Class 9 Science exams for motion-based questions?

To get full marks, always break down numerical problems by writing the given values, the relevant formula, and the final answer with correct units like $\text{m s}^{-1}$. For theoretical concepts like displacement versus distance, write point-wise differences and use clear examples.

### Is the free PDF for these Class 9 Science Chapter 4 solutions available for download?

Yes, SwaVid provides a complete and free PDF containing detailed step-by-step solutions for this chapter. You can easily access and download these verified resources directly from this page to aid your exam preparation for the 2026-27 session.

## Related pages

- [Class 9 Science chapters](https://www.swavid.com/science/class/9)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student