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Describing Motion Around Us

CBSE · Class 9 · Science

NCERT Solutions for Describing Motion Around Us — CBSE Class 9 Science.

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86 Questions Solved · 20 Sections

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Think It Over

1How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes?Show solution
A safe distance is the distance needed so that, if the truck ahead brakes suddenly, your vehicle can also stop in time. The chapter says this distance depends on the speed of the vehicle, because a vehicle moving faster needs a larger stopping distance.

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2Does this distance depend upon the speed with which we are moving?Show solution
Yes. The safe distance depends on the speed with which we are moving. A higher speed means the vehicle needs a greater distance to stop safely.

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Activity 4.1: Let us analyse

1As 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?Show solution
Yes. The ball moves along the same straight vertical line while going up and coming down, so it is a motion in a straight line.

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2For this motion, fill up the values in Table 4.1.Show solution
From the motion shown:

- At B, the ball has gone up to 80 cm from O.
- Total distance travelled = 80 cm
- Displacement = 80 cm upward
- At C, the ball has reached the top and starts coming back.
- Total distance travelled = 120 cm
- Displacement = 120 cm upward
- At the final O, it comes back to the start.
- Total distance travelled = 160 cm
- Displacement = 0 cm

So the missing entries are:

- B: 80 cm, 80 cm upward
- C: 120 cm, 120 cm upward
- O: 160 cm, 0 cm

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3Analyse the data filled in Table 4.1 and choose which of the following is true for displacement:Show solution
For displacement, the chapter says its magnitude is the distance between the object's positions at two instants, so it cannot exceed the total distance travelled. Therefore, the correct statement is (iii) Its magnitude is less than or equal to the total distance travelled.

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3(i)It is never zero.Show solution
Displacement can be zero if the object comes back to its starting point. So the statement is false.

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3(ii)Its magnitude can be greater than the total distance travelled.Show solution
The magnitude of displacement can never be greater than the total distance travelled. So this statement is false.

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3(iii)Its magnitude is less than or equal to the total distance travelled.Show solution
Yes. The magnitude of displacement is always less than or equal to the total distance travelled.

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3(iv)Its magnitude is less than the total distance travelled in all cases.Show solution
The magnitude of displacement is not less than the total distance travelled in all cases; it is less than or equal to it. So the statement is false.

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Pause and Ponder

1In 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?Show solution
The athlete’s displacement will be zero when she comes back to the starting point. In that case, the total distance travelled will be the entire path run by her on the track, which is not zero. For example, in Fig. 4.4 the athlete finally returns to the starting point, so displacement is zero while the total distance travelled is the full distance covered along the path.

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2Fuel used up in a vehicle depends on which of the following? Justify your answer.Show solution
Fuel used by a vehicle depends on the total distance travelled, not on displacement. Fuel consumption is related to how much the vehicle actually moves along its path. If a vehicle goes out and comes back, its displacement may be zero, but fuel is still used because the distance travelled is not zero.

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3A 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 itsShow solution
Yes, the ball rolling down an inclined track moves along a straight line path along the slope, so it is a straight line motion. But if we take the starting point O as the origin, we should not represent this motion by a horizontal line as in Fig. 4.3, because the motion is along the inclined track, not horizontally. The total distance travelled and the magnitude of displacement from O are equal at every position here, because the ball moves in one direction along the same straight line.

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4During a family road trip, you drive 200km200\mathrm{km} north in three hours. Afterwards, you drive 200km200\mathrm{km} south in two hours. Find the average speed and average velocity for your entire trip.Show solution
Total distance travelled =200+200=400= 200 + 200 = 400 km

Total time taken =3+2=5= 3 + 2 = 5 h

### Average speed
average speed=total distance travelledtime interval=4005=80 km/h \text{average speed} = \frac{\text{total distance travelled}}{\text{time interval}} = \frac{400}{5} = 80\ \text{km/h}

### Average velocity
The trip ends where it started, so displacement =0= 0 km.

average velocity=displacementtime interval=05=0 km/h \text{average velocity} = \frac{\text{displacement}}{\text{time interval}} = \frac{0}{5} = 0\ \text{km/h}

So the average speed is 80 km/h and the average velocity is 0 km/h.

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5(i)magnitude of average velocity of an object equal to its average speed?Show solution
The magnitude of average velocity is equal to average speed when the object moves in one direction only without turning back, so that distance travelled = magnitude of displacement.

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5(ii)magnitude of average velocity of an object zero while its average speed is not zero?Show solution
The magnitude of average velocity can be zero while average speed is not zero when the object returns to its starting point, so displacement is zero but distance travelled is not zero.

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4A 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?Show solution
Yes. The ball moves along the same straight vertical line while going up and coming back, so it is motion in a straight line.

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5Under what condition(s) is theShow solution
For motion in a straight line, the average speed and the magnitude of average velocity are equal if the object moves in one direction only, without turning back. Then the distance travelled is equal to the magnitude of displacement.

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Activity 4.2: Let us calculate

1The magnitude of average acceleration of cars is generally specified as the time taken by the car to go from 0kmh−10\mathrm{km}\mathrm{h}^{-1} to 100kmh−1100\mathrm{km}\mathrm{h}^{-1}. Look it up on the internet and find this time for various cars, and record those in Table 4.2.Show solution

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2Calculate the magnitude of average acceleration for each car.Show solution
The magnitude of average acceleration is given by
a=change in velocitytime interval a = \frac{\text{change in velocity}}{\text{time interval}}
For each car in the activity, first note the time taken to go from 00 to 100 km h−1100\ \text{km h}^{-1}, convert if needed, and then use the formula. Since the textbook asks students to look up different cars on the internet, there is no single fixed numerical answer from the chapter itself.

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Activity 4.3: Let us plot a graph

1Take a sheet of graph paper. This paper is pre-divided into small squares (Fig. 4.11a), making it easier to plot data accurately.Show solution

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2On 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.Show solution

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3Refer 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.Show solution

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4Determine 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. The scale can beShow solution
The table in the chapter specifies the scale as:

- X-axis: 5 divisions = 1 s
- Y-axis: 5 divisions = 20 m

So this is the correct choice.

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5Use the chosen scale to mark values for time (1 s, 2 s, ...) along the X-axis from the origin. Similarly, mark values for position (20 m, 40 m, ...) along the Y-axis (Fig. 4.11b).Show solution

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6(i)Table 4.3 shows that at time 0s0\mathrm{s}, the position is also 0m0\mathrm{m}. The point corresponding to this set of values on the graph will therefore be the origin itself.Show solution

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6(ii)At 1 s, the position of vehicle is at 20m20\mathrm{m}. To mark these values, look for the point that represents 1 s on the X-axis. Draw a line parallel to the Y-axis atShow solution

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6(iii)Similarly, plot on the graph paper all points corresponding to positions of the vehicle at different instants of time.Show solution

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7Once all points are plotted, connect them to create the position-time graph for the vehicle's motion (Fig. 4.11c). It is a straight line for the data given in Table 4.3.Show solution

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6Begin plotting points on the graph paper to represent each set of time and position values from Table 4.3.Show solution

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Activity 4.4: Let us calculate

1In 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.Show solution

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2Extend the horizontal line from A and a triangle ABC is formed. What do the sides BC and CA of the triangle represent? BC represents the change in position (s2−s1)(s_2 - s_1), and AC represents the change in time (t2−t1)(t_2 - t_1).Show solution
In the triangle formed on the position-time graph:

- BC represents the change in position, i.e. s2−s1s_2 - s_1
- CA represents the change in time, i.e. t2−t1t_2 - t_1

So the sides mean exactly those quantities.

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3As per Eq. (4.2a), by dividing the change in position (BC) by the change in time (CA), you get the average velocityShow solution

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4By extracting values of time t1t_1 and t2t_2, and distances s1s_1 and s2s_2 from the graph, the magnitude of average velocity can be calculated asShow solution
Using the graph values:

v=s2−s1t2−t1=80 m−40 m4 s−2 s v = \frac{s_2 - s_1}{t_2 - t_1} = \frac{80\,\text{m} - 40\,\text{m}}{4\,\text{s} - 2\,\text{s}}
=40 m2 s=20 m s−1 = \frac{40\,\text{m}}{2\,\text{s}} = 20\,\text{m s}^{-1}

So the magnitude of average velocity is 20 m s⁻¹.

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2Extend the horizontal line from A and a triangle ABC is formed. What do the sides BC and CA of the triangle represent?Show solution
In the triangle formed on the position-time graph:

- BC represents the change in position (s2−s1)(s_2 - s_1).
- CA represents the change in time (t2−t1)(t_2 - t_1).

Dividing BC by CA gives the average velocity.

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Example 4.6

1What does the graph shown in Fig. 4.15 indicate about the nature of motion of the vehicle?Show solution
The graph in Fig. 4.15 shows a straight line parallel to the time axis, which means the position of the vehicle is constant. So the vehicle is at rest at a fixed distance from the origin, here 40 m from the origin.

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Example 4.7

1The position-time graphs of two objects A and B are given in Fig. 4.16a. The magnitude of average velocity of which object is higher?Show solution
The object whose position-time graph has the steeper slope has the higher magnitude of average velocity. In Fig. 4.16a, object B has the steeper graph, so B has the higher average velocity.

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4.2.3 Velocity-time graphs

1What does the shape of the velocity-time graph indicate about the nature of motion?Show solution
The shape of a velocity-time graph tells the nature of motion:

- A horizontal straight line parallel to the time axis means constant velocity and zero acceleration.
- A straight rising line means velocity is increasing with constant acceleration.
- A straight falling line means velocity is decreasing with constant negative acceleration.

So the shape indicates whether the motion is uniform or accelerated.

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2Which physical quantities can be obtained from a velocity-time graph?Show solution
A velocity-time graph can be used to obtain:

- the velocity of the object at each instant of time,
- the acceleration from the slope of the graph,
- the displacement from the area enclosed by the graph and the time axis.

So, from a velocity-time graph we can find velocity, acceleration, and displacement.

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Example 4.8

1Suppose a car is moving on a highway and brakes are applied, which cause an acceleration of −4 m s−2-4\ \mathrm{m\ s^{-2}}. How much will be the distance travelled by the car before coming to a stop, if the car was moving with a velocity of (i) 54 km h−154\ \mathrm{km\ h^{-1}}, and (ii) 108 km h−1108\ \mathrm{km\ h^{-1}} when the brakes were applied?Show solution
Using the kinematic equation

v2=u2+2asv^2=u^2+2as

For stopping, final velocity v=0v=0 and acceleration a=−4 m s−2a=-4\,\mathrm{m\,s^{-2}}.

### (i) When u=54 km h−1u=54\,\mathrm{km\,h^{-1}}
Convert to m/s:

54×10003600=15 m s−154\times \frac{1000}{3600}=15\,\mathrm{m\,s^{-1}}

Now,

02=152+2(−4)s0^2=15^2+2(-4)s
0=225−8s0=225-8s
8s=2258s=225
s=2258=28.125 m≈28.1 ms=\frac{225}{8}=28.125\,\mathrm{m} \approx 28.1\,\mathrm{m}

### (ii) When u=108 km h−1u=108\,\mathrm{km\,h^{-1}}
Convert to m/s:

108×10003600=30 m s−1108\times \frac{1000}{3600}=30\,\mathrm{m\,s^{-1}}

Now,

02=302+2(−4)s0^2=30^2+2(-4)s
0=900−8s0=900-8s
8s=9008s=900
s=9008=112.5 ms=\frac{900}{8}=112.5\,\mathrm{m}

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2Suppose a car is moving on a highway and brakes are applied, which cause an acceleration of −4 m s−2-4\ \mathrm{m\ s^{-2}}. How much will be the distance travelled by the car before coming to a stop, if the car was moving with a velocity of (i) 54 km h−154\ \mathrm{km\ h^{-1}}, and (ii) 108 km h−1108\ \mathrm{km\ h^{-1}} when the brakes were applied?Show solution
The given values are the same as in the chapter’s Example 4.8.

Using

v2=u2+2asv^2=u^2+2as

with v=0v=0 and a=−4 m s−2a=-4\,\mathrm{m\,s^{-2}}:

### (i) u=54 km h−1=15 m s−1u=54\,\mathrm{km\,h^{-1}}=15\,\mathrm{m\,s^{-1}}

0=152+2(−4)s0=15^2+2(-4)s
0=225−8s0=225-8s
s=2258=28.125 m≈28.1 ms=\frac{225}{8}=28.125\,\mathrm{m} \approx 28.1\,\mathrm{m}

### (ii) u=108 km h−1=30 m s−1u=108\,\mathrm{km\,h^{-1}}=30\,\mathrm{m\,s^{-1}}

0=302+2(−4)s0=30^2+2(-4)s
0=900−8s0=900-8s
s=9008=112.5 ms=\frac{900}{8}=112.5\,\mathrm{m}

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4.4 Motion in a Plane

1Motion in a plane, such as a vehicle overtaking another, the path of a kicked ball or a satellite moving in a circular path, is called motion in two dimensions (Fig. 4.21).Show solution

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4.4.1 Uniform circular motion

1Do you remember learning about circular motion in an earlier grade? When an object moves in a circular path, its motion is called circular motion.Show solution
Yes. An object moving in a circular path is said to be in circular motion.

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2Suppose a child is sitting on a moving merry-go-around. The child, moving on a circular path, moves from A to B to C as shown in Fig. 4.22. What is the distance travelled by the child? What is their displacement from their original position?Show solution
The child travels along the curved path ABC. So the distance travelled is the length of the path from A to B to C.

The displacement is the straight-line distance from the original position to the final position, i.e. AC.

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3What is the distance travelled by the child in making one revolution (going round the circle once)?
4If an object takes time TT to make one revolution, its average speed vavv_{av} will be (using Eq. 4.1)
5while the average velocity during the time interval TT will be 0, since the displacement is 0.
6Let us now consider a particular case of circular motion where the speed of the object is constant. When an object moves in a circular path with constant (uniform) speed, its motion is called uniform circular motion.
7In case of uniform circular motion, the speed is constant but what about the direction of velocity at an instant? Is it changing?
8When an object moves in a circular path with constant (uniform) speed, its motion is called uniform circular motion.

Activity 4.5: Let us investigate

1Take a ring, such as an adhesive tape ring and one marble.
2Place 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).
3Predict what will happen if you lift the ring while the marble is moving.
4Now, 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?
5Repeat the activity multiple times to confirm the result.

Revise, Reflect, Refine

1My father went to a shop from home which is located at a distance of 250 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?
2A 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 m, find:
3A 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?
4A car starts from rest and its velocity reaches 24 m s⁻¹ in 6 s. Find the average acceleration and the distance travelled in these 6 s.
5A motorbike moving with initial velocity 28 m s⁻¹ and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.
6Fig. 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.
7A 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).
8A truck driver driving at the speed of 54 km h⁻¹ notices a road sign with a speed limit of 40 km h⁻¹ (Fig. 4.29) for trucks. He slows down to 36 km h⁻¹ in 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.
9A car starts from rest and accelerates uniformly to 20 m s⁻¹ in 5 seconds. It then travels at 20 m s⁻¹ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.
10A bus is travelling at 36 km h⁻¹ when the driver sees an obstacle 30 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 m s⁻². Will the bus be able to stop before reaching the obstacle?
11A 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.
12The velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist
13A 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.
14On entering a state highway, a car continues to move with a constant velocity of 6 m s⁻¹ for 2 minutes and then accelerates with a constant acceleration 1 m s⁻² for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.
15Two cars A and B start moving with a constant acceleration from rest, in a straight line. Car A attains a velocity of 5 m s⁻¹ in 5 s. Car B attains a velocity of 3 m s⁻¹ in 10 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).
16Rohan 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:
17Also, calculate the displacement and average acceleration in the 120 s time interval.
14On entering a state highway, a car continues to move with a constant velocity of 6 m s⁻¹ for 2 minutes and then accelerates with a constant acceleration 1 m s⁻¹ for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.

The Journey Beyond

1Take a cardboard disc (radius ~8 cm) (Fig. 4.33). Write numbers 1 to 12 on the outer part (7 cm from the centre) and the letters 'ABCDEF' on the inner part (4 cm from the centre), using the same font size. Spin the disc slowly, then faster, and observe how the numbers and letters appear. Why do the numbers fade or disappear while the letters remain visible? Are the speeds of the numbers and letters the same or different? Justify your answer.
2Many smartphones have an inbuilt accelerometer that can detect very small accelerations. Install an app, such as Phyphox (phyphox.org) and open 'Accelerometer (without g g )'. Note the readings when (i) the phone is on an outstretched palm, and (ii) the phone is kept on the floor. What differences do you observe? What does this tell you about motion and acceleration in real situations? (Such tiny, involuntary movements are also studied in medical research, for example, in movement disorders) This activity is recommended to be performed as a classroom group activity facilitated by teacher.
3For motion in a straight line with constant acceleration, we derived two primary equations given by Eq. (4.4a) and (4.4b). Using these two equations, three more equations can be derived, out of which we derived one given in Eq. (4.4c). Derive the remaining two equations given below
4Plot graphs for data given in Table 4.4, using different X and Y scales, on different graph papers. Compare the graphs to find how the appearance of graph is affected by the choice of scales and decide which scale is better and why. Now repeat this with any graph plotting app. Such apps generally automatically adjust the axes to fit the data well on the screen.
5Talk to a motor mechanic about how a vehicle's braking or stopping distance is affected by: (i) wet roads, (ii) worn-out tyres, (iii) higher vehicle mass, (iv) driving at night, (v) fog, (vi) severe weather (rain, snow, storm), and (vii) driver reaction time. Using this information, design safety posters for your school and prepare a short skit to present it in the assembly.
6Take a cardboard disc (radius ∼8\sim 8 cm) (Fig. 4.33). Write numbers 1 to 12 on the outer part (7 cm from the centre) and the letters 'ABCDEF' on the inner part (4 cm from the centre), using the same font size. Spin the disc slowly, then faster, and observe how the numbers and letters appear. Why do the numbers fade or disappear while the letters remain visible? Are the speeds of the numbers and letters the same or different? Justify your answer.
7Take a cardboard disc (radius ~ 8 cm) (Fig. 4.33). Write numbers 1 to 12 on the outer part (7 cm from the centre) and the letters 'ABCDEF' on the inner part (4 cm from the centre), using the same font size. Spin the disc slowly, then faster, and observe how the numbers and letters appear. Why do the numbers fade or disappear while the letters remain visible? Are the speeds of the numbers and letters the same or different? Justify your answer.

Example 4.5

1For a vehicle starting from rest and speeding up, the data for position and time are given in Table 4.4. Plot the position-time graph corresponding to it.

4.2.2 Position-time graphs

1What does the shape of the position-time graph indicate about the nature of motion?
2Which physical quantities can be obtained from a position-time graph?

Example 4.3: A bus is moving on a long straight highway (Fig. 4.9) with a velocity of 36 km h$^{-1}$. The driver presses the accelerator for a time interval of 10 s and velocity of the bus increases to 54 km h$^{-1}$. For some time, the bus moves at a constant velocity. Then, the driver notices an obstacle on the road ahead and presses the brake. The bus comes to a stop in a time interval of 5 s. Find the average acceleration in the two time intervals, (i) when the accelerator was pressed, and (ii) when the brakes were pressed.

1(i) When the driver presses the accelerator
2(ii) When the driver presses the brake

4.3 Kinematic Equations for Motion in a Straight Line with Constant Acceleration

1Using these equations, it is possible to predict position or velocity of the object at a future time.

Example 4.5: For a vehicle starting from rest and speeding up, the data for position and time are given in Table 4.4. Plot the position-time graph corresponding to it.

1Plot the position-time graph corresponding to it.

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