Understanding Motion — Long Answer Questions (Class 9 Science, Physics)
Medium Level (Application & Explanation)
Q1. Explain what a reference point is and why choosing a proper reference point is important when describing motion. Give two classroom examples.
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Answer:
- A referencepoint is a fixed location used to describe the position of another object. It acts as a basis for stating where something is.
- Choosing a proper point is important because it makes the description of motion clear, consistent, and comparable. Without a clearreferencepoint, statements like “the book moved” are ambiguous.reference
- A good point is usually stationary and easy to identify, such as a wall, the teacher’s desk, or the classroom door.reference
- Classroom examples: (1) “My desk is 3 m north of the teacher’s desk” uses the teacher’s desk as the point. (2) “The globe is 2 m east of the blackboard” uses the blackboard as thereferencepoint.reference
- In experiments, a fixed point helps measure displacement and notice small changes in position.reference
Q2. A student walks 10 km east, then 4 km west, then 2 km east. Calculate the distance and displacement. Explain the difference using this example.
Answer:
- Total distance is the actual path length: 10 km + 4 km + 2 km = 16 km. Distance is scalar and only adds path lengths.
- Displacement is the straight-line change from start to end with direction. Net eastward movement = 10 − 4 + 2 = 8 km east. Displacement is a vector and gives direction.
- This example shows that distance (16 km) is greater than displacement (8 km east) because the student backtracked partly.
- Distance depends on the actual path taken, while displacement depends only on the initial and final positions.
- If the student returned to the starting point, displacement would be zero even if distance were large.
Q3. How can you distinguish uniform and non-uniform motion from a distance–time graph? Describe what each graph looks like and what information you can read from it.
Answer:
- In a distance–time graph, the vertical axis shows distance and the horizontal axis shows time.
- Uniform motion appears as a straight line with a constant slope. The slope equals speed, so an unchanging slope means constant speed and zero acceleration.
- Non-uniform motion appears as a curved line or a line with changing slope. The slope (instantaneous speed) changes with time, indicating acceleration or deceleration.
- From the graph you can read distance at any time and, by measuring slope, determine speed. A horizontal line means the object is at rest. A steeper slope means a larger speed. A changing slope indicates the object’s speed is changing.
- Thus, graph shape tells whether motion is uniform or non-uniform and gives quantitative speed information.
Q4. A car moves at a steady speed of 60 km/h for 2 hours. Use the formula for uniform motion to find the distance covered and explain why the formula works here.
Answer:
- For uniform motion, distance = speed × time because speed remains constant over time.
- Given speed = 60 km/h and time = 2 h, distance = 60 × 2 = 120 km.
- The formula works because the car covers the same distance in each equal time interval. Multiplying the constant rate (speed) by total time gives the total path length.
- This assumes no change in speed and motion along the same direction. If speed varied, we could not use this simple relation and would need to integrate or add distances covered in subintervals.
- The result 120 km also matches practical expectation for long, steady highway travel.
Q5. Explain why a round trip can have a large distance but zero displacement, and state one real-life example where this distinction matters.
Answer:
- A round trip means an object returns to its starting point after travel. Distance measures the entire path length and remains positive, while displacement measures the net change in position and becomes zero because final and initial positions coincide.
- For example, walking 5 km around a park and coming back to the starting gate gives distance = 5 km but displacement = 0.
- This distinction matters in navigation and vector calculations. In physics problems involving net force or change in position, displacement is what determines direction-sensitive results. For example, in measuring how far a rescue team must be transported from one location to another, distance affects fuel use, while displacement determines final location relative to start.
- Hence, both quantities give different but important information.
High Complexity (Analytical & Scenario-Based)
Q6. A bus travels 10 km east in 0.5 h, then 6 km west in 0.3 h, and finally 4 km east in 0.2 h. Calculate the total distance, displacement, average speed, and average velocity. Explain the physical meaning of the results.
Answer:
- Total distance = 10 + 6 + 4 = 20 km. Distance is scalar, so we add all path lengths.
- Net displacement = eastward: 10 − 6 + 4 = 8 km east. This is a vector from start to end.
- Total time = 0.5 + 0.3 + 0.2 = 1.0 h.
- Average speed = total distance / total time = 20 / 1 = 20 km/h. This tells how fast the bus covered the whole path on average.
- Average velocity = displacement / total time = 8 / 1 = 8 km/h east. This tells how quickly and in which direction the bus moved from its start to end point.
- Physically, average speed is larger than average velocity when the path includes turns or backtracking; speed measures total motion while velocity measures net change of position.
Q7. A runner completes one full lap on a circular track at a constant speed. Discuss whether the motion is uniform or non-uniform, and state the runner’s distance, displacement, and acceleration during the lap.
Answer:
- The runner has constant speed, so the magnitudes of velocity are constant. However, velocity includes direction, which continuously changes around the circle. Because direction changes, velocity is not constant; therefore motion is non-uniform in the vector sense.
- Distance after one full lap equals the circumference of the track (a positive number). Displacement after one lap is zero because the final position coincides with the start.
- Even at constant speed, the runner experiences centripetal acceleration directed toward the track center; this acceleration changes the direction of velocity though not its magnitude.
- Thus, constant speed does not imply zero acceleration when the motion is along a curved path.
Q8. While sitting in a moving train you see trees, stations, and other trains. Explain how relative motion and reference points determine which objects you perceive as at rest or in motion. Give two examples.
Answer:
- Motion is observed relative to the chosen point. An object may appear at rest with respect to onereferencebut move relative to another.reference
- Example 1: With the train as your point, the seat beside you and the aisle are at rest since they do not change position relative to you. However, trees outside move backward because their position changes relative to you.reference
- Example 2: If you choose the ground as , the train is in motion. If another train on an adjacent track moves at the same speed and direction as yours, it may appear at rest relative to you, though both move relative to the ground.reference
- Choosing a clear, fixed simplifies description and avoids confusion in comparing motions.reference
Q9. A distance–time graph shows: from 0–2 s the distance increases linearly from 0 to 6 m; from 2–5 s the graph is horizontal at 6 m; from 5–7 s the distance decreases linearly from 6 m to 2 m. Describe the motion in each interval, compute the average speed for the whole 7 s, and state the displacement.
Answer:
- Interval 0–2 s: distance increases from 0 to 6 m, so the object moves forward at constant speed = slope = 6/2 = 3 m/s (uniform motion).
- Interval 2–5 s: distance constant at 6 m, so the object is at rest for 3 s.
- Interval 5–7 s: distance decreases from 6 to 2 m, meaning the object moves back toward the start with constant speed magnitude = (6−2)/2 = 2 m/s, direction reversed.
- Total distance traveled = forward 6 m + rest 0 + backward 4 m = 10 m. Total time = 7 s. Average speed = 10 / 7 ≈ 1.43 m/s.
- Displacement = final position − initial position = 2 − 0 = 2 m from start.
- These values show how distance and displacement differ when motion reverses.
Q10. In an experiment measuring small horizontal displacements of a toy car, a student places the reference point on the car itself instead of a fixed wall. Explain the likely errors in reported displacement and how to correct them. Discuss why a fixed reference is essential in measurements.
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Answer:
- If the is placed on the car, the measured position of the car relative to that movingreferencewill remain nearly constant, giving near-zero displacements, which is wrong. The reported data will not show the car’s actual movement relative to the lab.reference
- This error occurs because the must be stationary; a movingreferencehides true motion and ruins repeatability.reference
- To correct this, place the on a fixed object like a wall, table edge, or a marker anchored to the floor. Measure positions of the car relative to that fixedreferenceat different times. Use a ruler aligned to the fixedreferenceor attach a stationary scale to the track.reference
- A fixed ensures accurate, comparable, and meaningful displacement data and allows correct calculation of velocity and acceleration.reference