NCERT Solutions for Chapter 4: Describing Motion Around Us
Here are the NCERT solutions for Chapter 4: Describing Motion Around Us, with questions kept exactly as they appear in the sources and answers provided in simplified English.
Exercise Questions (Revise, Reflect, Refine - Page 68)
Question 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?
The father traveled the 250 m path between home and the shop four times: once to the shop, once back for the bag, once more to the shop, and a final time back home.
• Total distance: 250 + 250 + 250 + 250 = 1000 m.
• Displacement: Since he ended up exactly where he started (at home), his final change in position is 0 m.
Question 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.
Each floor is 3 m high.
• (i) Total distance: The student went up 4 floors (4 x 3 = 12 m) and then came down 2 floors (2 x 3 = 6 m). The total path traveled is 12 + 6 = 18 m.
• (ii) Displacement: The student ended up on the 2nd floor. Since they started on the ground floor, their final position is 2 floors above the start (2 x 3 = 6 m). The displacement is 6 m upwards.
Question 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?
Yes, it is possible. Even if the speed is constant, the scooter accelerates if it changes direction, such as when turning on a curved road. Velocity includes both speed and direction, so changing direction means the velocity is changing, which is called acceleration.
Question 4. A car starts from rest and its velocity reaches 24 ms⁻¹ in 6 s. Find the average acceleration and the distance travelled in these 6 s.
• Average acceleration: Acceleration = (Final speed - Initial speed) / Time. So, (24 - 0) / 6 = 4 m/s².
• Distance travelled: Using the formula s = ut + ½at², distance = (0 x 6) + ½ x 4 x 6². This equals 0 + 2 x 36 = 72 m.
Question 5. A motorbike moving with initial velocity 28 ms⁻¹ and constant acceleration stops after travelling 98 m. Find the acceleration of the motorbike and the time taken to come to a stop.
• Acceleration: Using the formula v² = u² + 2as, where final speed (v) is 0: 0 = 28² + 2 x a x 98. This gives a = -784 / 196 = -4 m/s². The negative sign means the bike is slowing down (deceleration).
• Time: Using v = u + at: 0 = 28 + (-4) x t. This gives t = 28 / 4 = 7 s.
Question 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.
No, they never have equal velocity. On this type of graph, velocity is shown by the steepness (slope) of the line. Since both lines are straight but have different steepness, Object A always moves at a higher constant speed than Object B. They meet at one point, but that only means they are at the same place, not that they are going the same speed.
Question 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 10s time interval is equal since they have the same initial and final positions. (ii) The average speeds of both over the 10s time interval are equal since both cover equal distances 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 1st time interval is greater than that of B since B's speed is lower than A's in some segments.
Options (i) and (ii) are correct. Both objects start and end at the same place, so their total displacement is zero, making their average velocities equal. They also cover the same total distance in the same time, so their average speeds are the same.
Question 8. A 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.
First, convert speeds to meters per second: 54 km/h = 15 m/s and 36 km/h = 10 m/s.
• Distance: Using average velocity, distance = [(Initial speed + Final speed) / 2] x Time. So, [(15 + 10) / 2] x 36 = 12.5 x 36 = 450 m.
Question 9. A car starts from rest and accelerates uniformly to 20 ms⁻¹ in 5 seconds. It then travels at 20 ms⁻¹ for 10 seconds and finally applies the brake (with uniform acceleration) to stop in 6 seconds. Find the total distance travelled.
• Phase 1 (Speeding up): Distance = [(0 + 20) / 2] x 5 = 50 m.
• Phase 2 (Constant speed): Distance = 20 x 10 = 200 m.
• Phase 3 (Slowing down): Distance = [(20 + 0) / 2] x 6 = 60 m.
• Total distance: 50 + 200 + 60 = 310 m.
Question 10. A 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 ms⁻². Will the bus be able to stop before reaching the obstacle?
Speed is 36 km/h, which is 10 m/s.
• Reaction distance: In 0.5 seconds of thinking, the bus moves 10 x 0.5 = 5 m.
• Braking distance: Using v² = u² + 2as: 0 = 10² + 2 x (-2.5) x s. This gives s = 100 / 5 = 20 m.
• Total stopping distance: 5 m + 20 m = 25 m.
• Conclusion: Since 25 m is less than 30 m, yes, the bus stops safely.
Question 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.
It depends on your frame of reference. To someone standing on Earth, the object is at rest because its position isn't changing relative to the ground. However, relative to the Sun, the object is moving because the Earth is spinning and traveling through space.
Question 12. The 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 (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.
• Shading: Shade the rectangle between 20 s and 100 s for constant velocity. Shade the slope going down between 100 s and 120 s for decreasing velocity.
• Total Displacement: Area 1 (Triangle) = 30 m; Area 2 (Rectangle) = 240 m; Area 3 (Trapezium) = 50 m. Total = 320 m.
• Average Acceleration: (Final speed - Initial speed) / Total time = (2 - 0) / 120 = 1/60 m/s².
Question 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.
By calculating the area under the different parts of the graph, the total estimated distance is approximately 49.15 km.
Question 14. On entering a state highway, a car continues to move with a constant velocity of 6 ms⁻¹ for 2 minutes and then accelerates with a constant acceleration 1ms⁻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.
• Constant speed part: In 120 seconds (2 mins), distance = 6 x 120 = 720 m.
• Acceleration part: Using s = ut + ½at²: (6 x 6) + ½ x 1 x 6² = 36 + 18 = 54 m.
• Total displacement: 720 + 54 = 774 m.
Question 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 ms⁻¹ in 5 s. Car B attains a velocity of 3 ms⁻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.
• Car A (in 5 s): Displacement is the area of the triangle = ½ x base x height = ½ x 5 x 5 = 12.5 m.
• Car B (in 10 s): Displacement = ½ x 10 x 3 = 15 m.
Question 16. Rohan studies science from 6 PM to 7:30 PM at home. Consider the tip of the minutes 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).
The hand is 7 cm long and rotates 1.5 times in 90 minutes.
• (i) Distance: 1.5 x circumference = 1.5 x 44 cm = 66 cm.
• (ii) Displacement: The hand starts at the 12 and ends at the 6. The straight-line distance is the diameter = 14 cm.
• (iii) Speed: Total distance / Total time = 66 cm / 5400 s = 11/900 cm/s.
• (iv) Velocity: Displacement / Total time = 14 cm / 5400 s = 7/2700 cm/s.
In-Text Questions
Question 1. How much distance should we maintain from the truck ahead to avoid a collision if it suddenly applies brakes?
You should maintain a safe distance, usually at least a 2-second gap, to have enough time to react and stop.
Question 2. Does this distance depend on the speed with which we are moving?
Yes. The faster you are moving, the more distance you need to come to a complete stop.
Question 3. Isn't motion in nature wonderful? But how do we study the wide variety of complex motions around us?
Yes, it is wonderful. We study it by observing natural examples like flowing water, flying birds, or moving clouds and using simple physics rules to describe them.
Question 4. How do we describe the position of an object?
We describe position by choosing a reference point and stating how far away the object is and in which direction.
Pause and Ponder Questions (Page 51-53)
Question 1. 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?
Displacement is zero when the athlete returns to the exact spot where they started. In this case, the total distance traveled will be a positive, non-zero number representing the entire path they ran.
Question 2. Fuel used up in a vehicle depends on which of the following? (i) Total distance travelled (ii) Displacement.
Fuel use depends on the total distance travelled. The engine has to work for every meter the car actually moves, regardless of the final displacement.
Question 3. A ball rolls down an inclined track... Is its motion a straight-line motion? Are the values of total distance travelled and magnitude of displacement from O equal or different...?
No, it is not straight-line motion because the track is curved. The total distance and displacement values are different at various points because the distance follows the curve of the track while displacement is a straight line.
Question 4. 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.
• Average speed: Total distance (400 km) / Total time (5 h) = 80 km/h.
• Average velocity: Since you returned to the start, total displacement is 0. So, average velocity = 0 / 5 = 0 km/h.
Question 5. 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?
• (i) They are equal when the object moves in a straight line without ever turning back.
• (ii) Average velocity is zero but speed is not when the object returns to its starting point.