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Exploration | Class 9 Science | Chapter 4 - Describing Motion Around Us | Q&A
CHAPTER 4

Describing Motion Around Us

Questions & Solutions | Exploration | Class 9 Science

Think It Over Page 68

Think It Over
How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies the brakes? Does this distance depend upon the speed with which we are moving?
Solution
The stopping distance depends on speed, reaction time, and braking capacity. Higher speeds require larger following distances to allow safe stopping.

Activity 4.1 Page 71

Activity 4.1
1. 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? 2. For this motion, fill up the values in Table 4.1. 3. Analyse the data filled in Table 4.1 and choose which of the following is true for displacement: (i) It is never zero. (ii) Its magnitude can be greater than the total distance travelled. (iii) Its magnitude is less than or equal to the total distance travelled. (iv) Its magnitude is less than the total distance travelled in all cases.
Solution
Yes, it is motion in a straight line (vertical). Option (iii) is correct: The magnitude of displacement is less than or equal to the total distance travelled.

Pause and Ponder

Page 71 (Q1)
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?
Solution
Displacement is zero when the athlete returns to the starting point. Total distance travelled will be the sum of all distances covered.
Page 71 (Q2)
Fuel used up in a vehicle depends on which of the following? (i) Total distance travelled (ii) Displacement
Solution
Fuel consumption depends on total distance travelled, not displacement, as energy is used for the entire path length.
Page 71 (Q3)
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?
Solution
Yes, it is straight line motion. Yes, it can be depicted on a horizontal line. Distance and displacement are equal because the ball moves in one direction without turning back.
Page 73 (Q4)
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
Total distance = 400 km, total time = 5 h. Average speed = 80 km/h. Displacement = 0 km, average velocity = 0 km/h.
Page 73 (Q5)
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?
Solution
(i) When the object moves in a straight line without changing direction. (ii) When the object returns to its starting point (zero displacement) but covers distance.

India's Scientific Contributions Page 72

India's Scientific Contributions
The concept that an object's speed is the distance travelled divided by the time taken is well-established, dating back to ancient times even in India, as seen in the treatise Aryabhatiya (5th century CE). The following problem, based on this concept, is from a comprehensive mathematical text, the Ganitakaumudi (14th century CE). Example 4.1: Consider two postmen. They start walking towards each other from a distance of 210 yojanas. One travels 9 yojanas per day and the other covers 5 yojanas per day. Can you determine in how many days they will meet each other?
Solution
Relative speed = 9 + 5 = 14 yojanas/day. Time = 210/14 = 15 days.

Ready to Go Beyond

Page 74
What we simply called 'velocity at an instant' is known as 'instantaneous velocity'. As the time interval around an instant is made progressively smaller, the change in average velocity gets smaller and smaller. When the time interval becomes infinitesimally small, the average value of velocity approaches a fixed value called the instantaneous velocity. You will learn more about this physical quantity in higher grades.
Solution
Instantaneous velocity is the velocity of an object at a specific instant of time, calculated as the limit of average velocity as time interval approaches zero.
Page 78
The intermediate points represent possible values for position of vehicle at intermediate times. They are correct if vehicle is moving with a constant speed.
Solution
For constant speed motion, the position-time graph is a straight line. Curved graphs indicate changing speed (acceleration).
Page 80
For the case of a curve also, the velocity at any instant can be calculated geometrically from the position-time graph. You will learn how to do it in higher grades.
Solution
Instantaneous velocity from a curved position-time graph is found by drawing a tangent at that point and calculating its slope.
Page 87
The velocity at a point is along the tangent to the circle at that point, in the direction of motion. A straight line that meets the circle at one and only point is called a tangent to the circle at that point. You will learn more about it in Mathematics.
Solution
In circular motion, the direction of velocity is always tangential to the circle at that point.

Activity 4.2 Page 75

Activity 4.2
1. The magnitude of average acceleration of cars is generally specified as the time taken by the car to go from 0 km/h to 100 km/h. Look it up on the internet and find this time for various cars, and record those in Table 4.2. 2. Calculate the magnitude of average acceleration for each car.
Solution
a = (v - u)/t. Convert 100 km/h to 27.78 m/s. For t = 10 s, a = 2.78 m/s². Answers vary by car model.

Activity 4.3 Page 77

Activity 4.3
1. Take a sheet of graph paper. 2. On the graph paper, draw two lines perpendicular to each other as shown in Fig. 4.11a. 3. Refer to Table 4.3. We will show time along the X-axis and position along the Y-axis. 4. Determine a suitable scale for each quantity. 5. Use the chosen scale to mark values for time and position. 6. Begin plotting points on the graph paper to represent each set of time and position values from Table 4.3. 7. Once all points are plotted, connect them to create the position-time graph.
Solution
The position-time graph for constant velocity motion is a straight line. Slope gives velocity.

Activity 4.4 Page 79

Activity 4.4
1. In the position-time graph we plotted (Fig. 4.11c), consider a part (say, AB) of the graph as shown in Fig. 4.14. 2. Extend the horizontal line from A and a triangle ABC is formed. 3. As per Eq. (4.2a), by dividing the change in position (BC) by the change in time (CA), you get the average velocity. 4. By extracting values from the graph, the magnitude of average velocity can be calculated.
Solution
Slope of position-time graph gives velocity. For points (2 s, 40 m) and (4 s, 80 m): v = (80-40)/(4-2) = 40/2 = 20 m/s.

Activity 4.5 Page 87

Activity 4.5
1. Take a ring, such as an adhesive tape ring and one marble. 2. 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). 3. Predict what will happen if you lift the ring while the marble is moving. 4. 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? 5. Repeat the activity multiple times to confirm the result.
Solution
When the ring is lifted, the marble moves in a straight line (tangent to the circle at the point of release) due to Newton's first law — an object in motion continues in a straight line unless acted upon by a force.

END-OF-CHAPTER EXERCISES Pages 88-90

1.
My 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?
Solution
Distance = 250 + 250 + 250 + 250 = 1000 m. Displacement = 0 m (returned to starting point).
2.
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 m, find: (i) the total vertical distance travelled, and (ii) their displacement from the starting point.
Solution
(i) Distance = ground to 4th (12 m) + 4th to 2nd (6 m) = 18 m. (ii) Displacement = ground to 2nd = 6 m upward.
3.
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
Yes, if the scooter is changing direction (e.g., moving in a circular path) while maintaining constant speed, it is accelerating due to change in direction.
4.
A car starts from rest and its velocity reaches 24 m/s in 6 s. Find the average acceleration and the distance travelled in these 6 s.
Solution
a = (24 - 0)/6 = 4 m/s². Distance = ut + ½at² = 0 + ½ × 4 × 36 = 72 m.
5.
A motorbike moving with initial velocity 28 m/s and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.
Solution
v² = u² + 2as → 0 = 784 + 2a×98 → a = -4 m/s². v = u + at → 0 = 28 - 4t → t = 7 s.
6.
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
If the slopes of the two curves are equal at any point (parallel tangents), they have equal velocity at that instant.
7.
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). (i) The average velocity of both over the 10 s time interval is equal since they have the same initial and final positions. (ii) The average speeds of both over the 10 s time interval are equal since both cover equal distance in equal time. (iii) The average speed of A over the 10 s time interval is lower than that of B since it covers a shorter distance than B in 10 seconds. (iv) The average speed of A over the 10 s time interval is greater than that of B since B's speed is lower than A's in some segments.
Solution
Option (iii) is correct if A covers shorter total distance than B.
8.
A truck driver driving at the speed of 54 km/h notices a road sign with a speed limit of 40 km/h (Fig. 4.29) for trucks. He slows down to 36 km/h in 36 s. What was the distance travelled by him during this time? Assume the acceleration to be constant while slowing down.
Solution
u = 54 km/h = 15 m/s, v = 36 km/h = 10 m/s, t = 36 s. s = (u+v)/2 × t = (25/2) × 36 = 12.5 × 36 = 450 m.
9.
A car starts from rest and accelerates uniformly to 20 m/s in 5 seconds. It then travels at 20 m/s for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.
Solution
Phase 1: s₁ = ½ × 20 × 5 = 50 m. Phase 2: s₂ = 20 × 10 = 200 m. Phase 3: s₃ = ½ × 20 × 6 = 60 m. Total = 310 m.
10.
A bus is travelling at 36 km/h 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/s². Will the bus be able to stop before reaching the obstacle?
Solution
u = 36 km/h = 10 m/s. Reaction distance = 10 × 0.5 = 5 m. Braking distance: v² = u² + 2as → 0 = 100 - 2×2.5×s → s = 20 m. Total = 5 + 20 = 25 m < 30 m. Yes, it will stop before the obstacle.
11.
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
Rest is relative. An object on Earth is at rest relative to Earth but in motion relative to the Sun or other celestial bodies.
12.
The velocity-time graph from 0s to 120 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 s time interval.
Solution
Displacement = area under v-t graph. Average acceleration = (v_f - v_i)/t = (0 - 0)/120 = 0 m/s².
13.
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
Distance = area under velocity-time graph. Calculate area of each geometric shape (rectangles, triangles) to estimate total distance.
14.
On entering a state highway, a car continues to move with a constant velocity of 6 m/s for 2 minutes and then accelerates with a constant acceleration 1 m/s² 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.
Solution
s₁ = 6 × 120 = 720 m. v after acceleration = 6 + 1×6 = 12 m/s. s₂ = 6×6 + ½×1×36 = 36 + 18 = 54 m. Total = 774 m.
15.
Two 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 in 5 s. Car B attains a velocity of 3 m/s 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.
Solution
Car A: a = 1 m/s², displacement = ½ × 5 × 5 = 12.5 m. Car B: a = 0.3 m/s², displacement = ½ × 3 × 10 = 15 m.
16.
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 cm (Fig.4.32).
Solution
Time = 90 minutes = 1.5 hours. Minute hand completes 1.5 revolutions. (i) Distance = 1.5 × 2πr = 1.5 × 2 × 22/7 × 7 = 66 cm. (ii) Displacement = 2r = 14 cm (opposite direction). (iii) Speed = 66/5400 = 0.0122 cm/s. (iv) Velocity = 14/5400 = 0.0026 cm/s.

Chapter Summary

  • Position describes the location of an object relative to a reference point.
  • Displacement is the net change in position (shortest distance between initial and final points).
  • Average speed = total distance / time; average velocity = displacement / time.
  • Average acceleration = change in velocity / time.
  • Kinematic equations for constant acceleration: v = u + at, s = ut + ½at², v² = u² + 2as.
  • Position-time graph slope gives velocity; velocity-time graph slope gives acceleration; area under v-t graph gives displacement.
  • Uniform circular motion has constant speed but changing direction, hence acceleration.

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Chapter 4: Describing Motion Around Us — All Questions with Solutions

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