Class 9 Science Chapter 9 Motion and Time: Complete Important Questions & Answers for 2025-26 CBSE Board

Motion and Time appears as a guaranteed 8–10 mark section in Class 9 Science final exams. Under the new rationalized syllabus, examiners focus heavily on: **Conceptual Clarity over Rote Definitions**: Instead of just defining speed, you're expected to compare uniform and non-uniform motion with real examples (a train on straight rails vs. a car in traffic). Questions on distance-time graphs now require you to extract velocity, acceleration direction, and even compare motion of two bodies on the same graph. **Application of Formulas**: Speed = Distance ÷ Time isn't just memorized—it's applied to multi-step scenarios. For example, if a person walks 120 m in 30 seconds, then runs the next 120 m in 15 seconds, you must calculate average speed and identify motion type. **Simple Pendulum Insights**: This is a 2–3 mark favorite because it bridges motion concepts with periodicity. You're expected to know T = 2π√(L/g) and interpret how length affects time period, a direct link to real-world timekeeping. **Graph Interpretation**: Distance-time and speed-time graphs now appear in 3–5 mark questions requiring you to find displacement, velocity, and even identify acceleration phases. By mastering these 18 questions, you'll be ready for any variation the examiner poses—whether it's a numerical on pendulum frequency or a graph-reading task.

**Question 1**: A car travels 200 m in 10 seconds. Its speed is: (a) 20 m/s (b) 2 m/s (c) 10 m/s (d) 0.5 m/s **Answer: (a) 20 m/s** Speed = Distance ÷ Time = 200 ÷ 10 = 20 m/s. This is straightforward NCERT formula application. --- **Question 2**: The SI unit of speed is: (a) km/h (b) m/s (c) cm/s (d) m/min **Answer: (b) m/s** The international standard unit for speed is metres per second. While km/h is commonly used in everyday life, m/s is the SI unit. --- **Question 3**: Which of the following is an example of non-uniform motion? (a) A ball rolling on smooth, flat ice (b) A cyclist moving down a steep hill, gradually accelerating (c) A planet orbiting the sun at constant speed (d) An electric train at constant velocity on a straight track **Answer: (b) A cyclist moving down a steep hill, gradually accelerating** Non-uniform motion means velocity changes. A cyclist accelerating has changing speed, so it's non-uniform. The others maintain constant speed. --- **Question 4**: A distance-time graph is a straight line parallel to the time axis. This represents: (a) Uniform motion (b) Non-uniform motion (c) Stationary object (d) Accelerated motion **Answer: (c) Stationary object** A horizontal line (parallel to time axis) means distance does not change over time—the object is at rest. No motion occurs. --- **Question 5**: The time period of a simple pendulum depends on: (a) The mass of the bob (b) The amplitude of oscillation (c) The length of the string (d) The colour of the bob **Answer: (c) The length of the string** From the formula T = 2π√(L/g), time period T depends only on length L and gravitational acceleration g. Mass, amplitude (for small angles), and colour have no effect.

**Question 1**: Define uniform and non-uniform motion with one example of each. **Answer**: *Uniform Motion*: Motion in which an object covers equal distances in equal intervals of time, maintaining constant speed and velocity. Example: A train moving at a steady 60 km/h on a straight, level track. *Non-Uniform Motion*: Motion in which an object covers unequal distances in equal intervals of time, so speed or direction (or both) changes. Example: A car accelerating from rest at traffic lights—it covers 5 m in the first 2 seconds but 15 m in the next 2 seconds. --- **Question 2**: A person walks 4 km in 1 hour, then cycles 8 km in 30 minutes. Calculate average speed for the entire journey. **Answer**: Total distance = 4 + 8 = 12 km Total time = 1 hour + 0.5 hours = 1.5 hours Average speed = Total distance ÷ Total time = 12 ÷ 1.5 = 8 km/h Alternatively, in SI units: 12 km = 12,000 m; 1.5 hours = 5,400 s Average speed ≈ 2.22 m/s --- **Question 3**: What does the slope (gradient) of a distance-time graph represent? If the slope is steep, what does it indicate? **Answer**: The slope of a distance-time graph represents speed (or velocity if direction is considered). Formula: Slope = Δdistance ÷ Δtime = Speed If the slope is *steep*, it means distance changes rapidly with time, indicating *high speed*. If the slope is *gentle*, the object moves *slowly*. A *horizontal line* (zero slope) means the object is stationary. --- **Question 4**: A simple pendulum with length 1 m completes 20 oscillations in 40 seconds. Find the time period and frequency. **Answer**: Time period T = Total time ÷ Number of oscillations = 40 ÷ 20 = 2 seconds Frequency f = 1 ÷ Time period = 1 ÷ 2 = 0.5 Hz (or 0.5 oscillations per second) Note: The NCERT formula T = 2π√(L/g) ≈ 2π√(1/10) ≈ 2 seconds confirms this result (g ≈ 10 m/s²). --- **Question 5**: Draw and label a distance-time graph for an object at rest for 5 seconds, then moving with uniform motion for the next 5 seconds. **Answer**: The graph has two segments: 1. From t = 0 to t = 5 s: A *horizontal line at a fixed distance* (e.g., d = 0), representing no motion. 2. From t = 5 to t = 10 s: A *straight line with positive slope*, rising from d = 0 to d = 50 m (for example), representing uniform motion at constant speed = 50 ÷ 5 = 10 m/s. The y-axis is labeled 'Distance (m)', x-axis is 'Time (s)'. The first segment is flat; the second is inclined upward at a constant angle.

**Question 1**: A train accelerates from rest. In the first second it covers 2 m, in the second second it covers 4 m, and in the third second it covers 6 m. Is this uniform or non-uniform motion? Calculate the average speed over 3 seconds. **Answer**: *Motion Type*: This is non-uniform motion because the train covers different distances (2 m, 4 m, 6 m) in equal time intervals (1 second each). The distances are increasing, indicating acceleration. *Average Speed Calculation*: Total distance = 2 + 4 + 6 = 12 m Total time = 3 s Average speed = 12 ÷ 3 = 4 m/s --- **Question 2**: From the following distance-time graph data, determine: (i) Which object is moving faster, A or B? (ii) The speed of object A. (iii) At what time do both objects meet? Assumed data: Object A travels 100 m in 10 s; Object B travels 100 m in 20 s. **Answer**: (i) *Faster object*: Object A is faster because it covers the same distance (100 m) in less time (10 s vs. 20 s). (ii) *Speed of A*: Speed = 100 ÷ 10 = 10 m/s Speed of B: Speed = 100 ÷ 20 = 5 m/s (iii) *Meeting point*: If both start from the same point at t = 0, they meet when distance is equal. At t = 20 s, both have traveled 100 m (A: 20 × 10 = 200 m... correction: they don't meet on a linear graph unless B starts ahead or A stops). On standard NCERT graphs, if both travel equal distances, they meet at the intersection point on the graph. --- **Question 3**: A simple pendulum has a length of 2.5 m. Calculate its time period (using g = 10 m/s² and π ≈ 3.14). How many complete oscillations will it make in 31.4 seconds? **Answer**: *Time Period*: T = 2π√(L/g) = 2 × 3.14 × √(2.5/10) = 6.28 × √(0.25) = 6.28 × 0.5 = 3.14 seconds *Number of Oscillations*: Oscillations = Total time ÷ Time period = 31.4 ÷ 3.14 = 10 complete oscillations --- **Question 4**: A bus travels 60 km in 2 hours on a bumpy road (non-uniform motion). Another bus travels 60 km in 3 hours on a smooth highway (uniform motion). Which bus has higher average speed? Will the uniform-motion bus ever be faster instantaneously during their journey? Explain. **Answer**: *Average Speeds*: Bus 1 (bumpy): 60 ÷ 2 = 30 km/h Bus 2 (smooth): 60 ÷ 3 = 20 km/h Bus 1 has higher average speed (30 km/h). *Instantaneous Speed*: Yes, the uniform-motion bus (Bus 2) maintains a constant 20 km/h throughout. The bumpy-road bus (Bus 1) has non-uniform motion, meaning its instantaneous speed varies. At certain moments, Bus 1 might move at 15 km/h, while at others it's 40 km/h. When Bus 1's instantaneous speed drops below 20 km/h, Bus 2 is faster instantaneously, even though Bus 1's average is higher. This shows average and instantaneous speeds are different concepts.

**Question 1**: A boy runs along a straight path. For the first 10 seconds, he covers 50 m at constant speed. Then he accelerates and covers the next 100 m in 10 seconds. Finally, he decelerates and covers 30 m in 10 seconds before stopping. (i) Calculate the speed in each segment. (ii) Classify each segment as uniform or non-uniform motion. (iii) Draw a distance-time graph for the entire motion. (iv) Calculate the overall average speed. **Answer**: (i) *Speed in Each Segment*: - Segment 1 (0–10 s): Speed₁ = 50 ÷ 10 = 5 m/s - Segment 2 (10–20 s): Speed₂ = 100 ÷ 10 = 10 m/s - Segment 3 (20–30 s): Speed₃ = 30 ÷ 10 = 3 m/s (ii) *Motion Classification*: - Segment 1: Uniform motion (constant speed = 5 m/s) - Segment 2: Non-uniform motion (acceleration; speed increased from 5 m/s to 10 m/s) - Segment 3: Non-uniform motion (deceleration; speed decreased from 10 m/s to 0 m/s) (iii) *Distance-Time Graph*: The graph has three segments: - (0, 0) to (10, 50): Straight line with slope 5 m/s (gentle incline) - (10, 50) to (20, 150): Straight line with slope 10 m/s (steeper incline) - (20, 150) to (30, 180): Straight line with slope 3 m/s (gentler incline than segment 1) The slope visually shows speed differences: steeper = faster. (iv) *Overall Average Speed*: Total distance = 50 + 100 + 30 = 180 m Total time = 10 + 10 + 10 = 30 s Average speed = 180 ÷ 30 = 6 m/s Note: Average speed (6 m/s) differs from the average of individual speeds (5 + 10 + 3)/3 = 6 m/s (coincidence here). Average speed is always total distance ÷ total time. --- **Question 2**: Two buses, A and B, start from the same point. Bus A moves at a constant speed of 40 km/h, while Bus B accelerates uniformly from rest, reaching 60 km/h in 2 hours. (i) Calculate Bus B's acceleration. (ii) After 2 hours, how far has each bus traveled? Which is ahead? (iii) Will Bus B ever catch up and overtake Bus A? Justify with calculations. **Answer**: (i) *Bus B's Acceleration*: Acceleration a = (Final velocity − Initial velocity) ÷ Time a = (60 − 0) ÷ 2 = 30 km/h per hour = 30 km/h² In m/s²: 30 km/h² = 30 ÷ (3.6)² ≈ 2.3 m/s² (for reference) (ii) *Distance After 2 Hours*: Bus A (uniform motion): Distance = Speed × Time = 40 × 2 = 80 km Bus B (uniform acceleration): Distance = (Initial velocity × Time) + (0.5 × a × t²) = (0 × 2) + (0.5 × 30 × 2²) = 0 + (0.5 × 30 × 4) = 60 km Bus A is ahead by 80 − 60 = 20 km. (iii) *Will Bus B Overtake?* Yes, Bus B will eventually overtake Bus A because: - Bus A maintains constant 40 km/h; its position is d_A = 40t - Bus B accelerates at 30 km/h²; its position is d_B = 15t² When d_B = d_A: 15t² = 40t 15t² − 40t = 0 t(15t − 40) = 0 t = 0 or t = 40/15 ≈ 2.67 hours At t ≈ 2.67 hours, both are at the same position. After this time, 15t² > 40t, so Bus B pulls ahead. By t = 3 hours: Bus A: 40 × 3 = 120 km Bus B: 15 × 3² = 135 km Bus B is now ahead. --- **Question 3**: A simple pendulum on Earth has a length of 0.4 m and a time period of 1.28 seconds. Another identical pendulum is taken to a planet where its time period becomes 1.6 seconds. Calculate the gravitational acceleration on that planet. **Answer**: *Given Data*: - Length L = 0.4 m (same for both) - On Earth: T_E = 1.28 s - On planet: T_P = 1.6 s - g_E = 10 m/s² (assumed) *Using the formula T = 2π√(L/g)*: On Earth: T_E = 2π√(L/g_E) 1.28 = 2π√(0.4/g_E) 1.28² = (2π)² × (0.4/g_E) 1.6384 = 39.48 × (0.4/g_E) g_E = (39.48 × 0.4) ÷ 1.6384 ≈ 9.65 m/s² ≈ 10 m/s² ✓ On the planet: T_P = 2π√(L/g_P) 1.6 = 2π√(0.4/g_P) 1.6² = (2π)² × (0.4/g_P) 2.56 = 39.48 × (0.4/g_P) g_P = (39.48 × 0.4) ÷ 2.56 g_P = 15.792 ÷ 2.56 g_P ≈ 6.17 m/s² *Interpretation*: The gravitational acceleration on that planet is approximately 6.17 m/s², which is lower than Earth's 10 m/s². This explains why the pendulum takes longer to oscillate—weaker gravity means slower motion.

**Case Study**: A fitness tracker records the running motion of an athlete over 30 seconds. The device shows: - 0–10 s: 80 m covered - 10–20 s: 120 m covered - 20–30 s: 60 m covered The athlete claims her motion was "uniform" and "fast." Analyze her claim and answer: (i) Is her motion truly uniform? Support with calculations. (ii) Calculate speed in each segment and identify the motion type for each. (iii) Sketch a distance-time graph. From the graph, determine when she was fastest and slowest. (iv) The athlete wants to maintain *uniform* motion at an average speed. What should her speed be, and how far would she cover in 45 seconds at this uniform speed? (v) **HOTS**: If the athlete's acceleration in segment 2 (0–10 s to 10–20 s transition) is constant, calculate this acceleration. Could she have maintained this acceleration for the entire 30 seconds? Why or why not? **Solution**: (i) *Uniformity Check*: Segment 1 (0–10 s): Speed = 80 ÷ 10 = 8 m/s Segment 2 (10–20 s): Speed = 120 ÷ 10 = 12 m/s Segment 3 (20–30 s): Speed = 60 ÷ 10 = 6 m/s Speeds are 8, 12, and 6 m/s—*not equal*. Motion is **non-uniform**, not uniform. Her claim is false. (ii) *Motion Classification*: - Segment 1: Uniform motion at 8 m/s - Segment 2: Uniform motion at 12 m/s (but overall motion is non-uniform as speed changes between segments) - Segment 3: Uniform motion at 6 m/s Each segment individually is uniform, but collectively, the motion is non-uniform. (iii) *Distance-Time Graph*: Plot points (0,0), (10,80), (20,200), (30,260). - Line from (0,0) to (10,80): slope = 8 m/s - Line from (10,80) to (20,200): slope = 12 m/s (steepest—fastest) - Line from (20,200) to (30,260): slope = 6 m/s (gentlest—slowest) **Fastest segment**: 10–20 s (12 m/s; steepest slope) **Slowest segment**: 20–30 s (6 m/s; gentlest slope) (iv) *Average Speed for Uniform Motion*: Total distance = 80 + 120 + 60 = 260 m Total time = 30 s Average speed = 260 ÷ 30 ≈ 8.67 m/s Distance in 45 seconds at uniform 8.67 m/s: Distance = 8.67 × 45 ≈ 390 m (v) *Acceleration (HOTS)*: Acceleration is the change in velocity over time. From segment 1 to segment 2 (assuming instantaneous transition at t = 10 s): Δv = 12 − 8 = 4 m/s Δt = (infinitesimal, but for practical calculation over 1 second) ≈ very high or undefined If we interpret "constant acceleration" as the change between average velocities: Acceleration ≈ (12 − 8) ÷ 10 = 0.4 m/s per 10 seconds = 0.04 m/s² *Could she maintain this for 30 seconds?* If a = 0.04 m/s², starting at v₀ = 8 m/s: At t = 30 s: v = 8 + (0.04 × 30) = 8 + 1.2 = 9.2 m/s Distance = v₀t + 0.5at² = 8(30) + 0.5(0.04)(30)² = 240 + 18 = 258 m But the athlete covered 260 m. The slight difference suggests constant acceleration doesn't fully explain her motion pattern. **Why?** Her segment 3 shows *deceleration* (6 m/s from 12 m/s), which constant positive acceleration cannot account for. She likely changed her effort mid-run, making acceleration non-constant. Real-world fitness involves varying intensity, not pure physics constants!

CBSETUTOR.ai uses adaptive AI to drill exactly these Motion and Time patterns every day. Here's how: **Daily Drill System**: Each morning, the AI assesses your weak areas (e.g., graph interpretation, pendulum formulas) and serves 2–3 questions tailored to that skill. You solve; the AI instantly provides step-by-step feedback, not just a mark. **Multi-Difficulty Progression**: You start with 1-mark conceptual MCQs ("What is speed?"), then progress to 3-mark numerical problems, and finally tackle 5-mark case studies. The AI adjusts difficulty based on your accuracy, ensuring you're always challenged but not overwhelmed. **Graph Interactivity**: Instead of static diagrams, our AI lets you *draw* and *manipulate* distance-time graphs. Plot a point, and it calculates slope instantly. Misread a graph? The AI shows you where your error was using visual overlays. **Pendulum Simulations**: Visualize how length affects time period. The AI runs a live simulation: change L in T = 2π√(L/g), and watch the pendulum's oscillation speed update in real-time. This cement's conceptual understanding faster than any textbook. **Board-Pattern Mimicry**: CBSETUTOR.ai's question bank mirrors the exact distribution of 2025-26 CBSE papers: 30% MCQs, 35% 2–3 mark questions, 25% 5-mark answers, 10% HOTS. You practice under real exam conditions, boosting confidence. **Instant Doubt Clarification**: Stuck on a calculation or concept? Ask the AI. It explains in plain English (not textbook jargon), with worked examples. "Why does a steeper graph mean faster motion?" → Instant visual + numerical answer. **Performance Analytics**: The AI tracks your progress. After 2 weeks of daily drills, you'll see which topics you've mastered (e.g., speed calculations) and which need more work (e.g., pendulum period). Parents can also see this in a dashboard. **Start a 3-day free trial at cbsetutor.ai**—no credit card needed—and experience personalized motion and time drills designed for Class 9 CBSE success.