Measuring the Rate of Motion – Long Answer Questions (Class 9 Physics)
Medium Level (Application & Explanation)
Q1. A student travels from home to school 6 km away. He cycles 3 km at 12 km/h and walks the remaining 3 km at 6 km/h. Calculate his average speed for the whole journey and explain why average speed is not simply the arithmetic mean of the two speeds.
Answer:
- Calculate time for each part: time = distance / speed.
- Cycling: t₁ = 3 km ÷ 12 km/h = 0.25 h.
- Walking: t₂ = 3 km ÷ 6 km/h = 0.5 h.
- Total distance = 6 km. Total time = t₁ + t₂ = 0.75 h.
- Average speed = Total distance / Total time = 6 km ÷ 0.75 h = 8 km/h.
- Explanation: The average speed depends on total distance and total time, not on simply averaging speeds. The arithmetic mean (12+6)/2 = 9 km/h assumes equal time at both speeds, which is not true here. Because the cyclist spends less time at 12 km/h and more time at 6 km/h, the weighted effect reduces the overall average to 8 km/h. Emphasize that average speed is a time-weighted quantity.
Q2. Explain the difference between speed and velocity using a simple example of walking to a park and returning home. Describe what happens to average speed and average velocity in this round trip.
Answer:
- Speed is the rate of change of distance and is a scalar (only magnitude). Velocity is the rate of change of displacement and is a vector (magnitude + direction).
- Example: Walk 500 m to the park in 10 minutes and walk back 500 m in 15 minutes. Total distance = 1000 m; total time = 25 minutes. Average speed = 1000 m ÷ 1500 s ≈ 0.67 m/s.
- Displacement for the round trip = 0 m, so average velocity = 0 m ÷ 1500 s = 0 m/s.
- Thus, even though you were moving (non-zero speed), your average velocity is zero because net displacement is zero. This shows how direction matters for velocity but not for speed.
Q3. Convert speeds between units: A vehicle moves at 90 km/h. Convert this speed into m/s. Explain why we convert units and give the conversion factor.
Answer:
- Conversion factor: 1 km = 1000 m and 1 hour = 3600 seconds. Thus 1 km/h = (1000 m) / (3600 s) = 5/18 m/s.
- Convert 90 km/h: 90 × (5/18) m/s = 90 × 0.2777... m/s = 25 m/s.
- Reason: We convert units to use the SI unit (m/s) in calculations, to compare speeds correctly, and to apply formulas that require consistent units (e.g., equations of motion). Using consistent units avoids numerical errors and makes physical meaning clear. Always state units with numbers to avoid confusion.
Q4. A distance–time graph shows a straight line with constant slope for a car, then a horizontal line, then a line with a smaller slope. Interpret each segment and explain how these relate to speed and uniform motion.
Answer:
- First straight line with constant slope: The car is moving with uniform speed. Constant slope on a distance–time graph means distance increases linearly with time, so speed (rise/run) is constant.
- Horizontal line: Distance does not change with time; the car is stationary (speed = 0).
- Second line with smaller slope: The car moves again but at a lower uniform speed than before (slope smaller means smaller speed).
- Relation to uniform motion: Each straight-line segment indicates uniform motion (constant speed). Changes in slope show changes in speed. The steeper the slope, the greater the speed.
Q5. Can an object have constant speed but changing velocity? Give an example and explain the physical reason behind the change in velocity.
Answer:
- Yes. Example: A car moving at constant speed around a circular roundabout. Even though its speed remains unchanged, its direction continuously changes.
- Velocity includes direction, so any change in direction means velocity changes. In circular motion, the velocity vector is always tangent to the path and rotates as the object moves.
- This change in velocity implies acceleration (called centripetal acceleration) directed towards the center of the circle. The acceleration is present even if speed is constant because acceleration measures rate of change of velocity (both magnitude and direction).
- Emphasize: constant speed ≠ zero acceleration if direction changes.
High Complexity (Analytical & Scenario-Based)
Q6. A boat needs to cross a river 200 m wide, flowing east at 3 m/s. The boat can row at 4 m/s relative to water and points directly north. Determine (a) the resultant velocity of the boat relative to ground (magnitude and direction), and (b) time taken to cross. Explain the vector reasoning in simple terms.
Answer:
- Treat velocities as perpendicular vectors: northward boat speed (relative to water) = 4 m/s; eastward river current = 3 m/s.
- (a) Resultant speed magnitude = √(4² + 3²) = √(16 + 9) = √25 = 5 m/s. Direction: angle θ east of north = arctan(3/4) ≈ 36.9°. So the boat moves 5 m/s in a direction 36.9° east of north relative to ground.
- (b) Time to cross depends only on the northward component of velocity because river width is north–south distance = 200 m. Northward speed relative to ground = 4 m/s. Time = distance / speed = 200 m ÷ 4 m/s = 50 s.
- Vector reasoning: Combine perpendicular components using Pythagoras for magnitude and arctangent for direction. Eastward drift does not change crossing time, only landing position downstream.
Q7. A runner completes three laps of 400 m each on a straight track: first lap at 5 m/s, second at 8 m/s, third at 6 m/s. Calculate the average speed and average velocity for the entire run. Explain why average velocity is different from average speed.
Answer:
- Total distance = 3 × 400 m = 1200 m.
- Time for each lap: t₁ = 400 ÷ 5 = 80 s; t₂ = 400 ÷ 8 = 50 s; t₃ = 400 ÷ 6 ≈ 66.67 s. Total time = 196.67 s.
- Average speed = Total distance / Total time = 1200 m ÷ 196.67 s ≈ 6.10 m/s.
- For average velocity, we need displacement. If the runner returns to the starting point after three laps on a straight track, displacement = 0. Average velocity = 0 ÷ 196.67 s = 0 m/s.
- Difference: Average speed uses total distance and is positive; average velocity uses net displacement and can be zero if start and end points coincide. Direction matters for velocity but not for speed.
Q8. Two towns A and B are 150 km apart. A car travels from A to B at 50 km/h and returns from B to A at 75 km/h. Compute the average speed for the full journey and explain why it cannot be found by simple averaging of speeds.
Answer:
- Distance one-way = 150 km. Round trip distance = 300 km.
- Time A→B = 150 ÷ 50 = 3 h. Time B→A = 150 ÷ 75 = 2 h. Total time = 5 h.
- Average speed = Total distance / Total time = 300 km ÷ 5 h = 60 km/h.
- Why not arithmetic mean? The arithmetic mean (50+75)/2 = 62.5 km/h assumes equal times at both speeds, which is false here. Average speed depends on total time spent at each speed. Because the car spends more time at the lower speed (50 km/h), the overall speed is pulled lower than the simple mean in a time-weighted sense. Emphasize that average speed is distance/time based and not a plain average of individual speeds.
Q9. A satellite moves in circular orbit around Earth at constant speed. Explain why its velocity changes continuously and describe qualitatively the acceleration acting on the satellite. What is the direction of this acceleration?
Answer:
- Even when speed is constant, the satellite’s velocity changes because its direction continuously changes as it moves around the circle. Velocity is a vector (magnitude + direction), so a changing direction means changing velocity.
- This change in velocity implies an acceleration. In circular motion with constant speed, this acceleration is called centripetal acceleration. It acts toward the centre of the circular path (towards the centre of Earth for the satellite).
- The centripetal acceleration keeps the satellite in orbit by constantly changing the velocity direction without altering speed. Its magnitude is v²/r, where v is speed and r is orbit radius. Emphasize inward direction and that acceleration is perpendicular to instantaneous velocity.
Q10. Design a simple classroom experiment to measure the speed of a toy car moving in a straight line using a metre scale and a stopwatch. Describe the procedure, calculations, and at least three precautions to reduce measurement errors.
Answer:
- Procedure: Mark a straight track with measured start and end points, say 2.00 m apart using a metre scale. Place the toy car at the start. Use a stopwatch to measure time for the car to travel from start to end. Repeat the run at least five times and record times t₁, t₂,...t₅.
- Calculation: For each run, speed = distance / time = 2.00 m ÷ t_i. Compute the average of these speeds (or average time first then compute speed) to obtain a more reliable value. Report speed in m/s.
- Precautions to reduce errors:
- Start and stop timing precisely (use the same student or an electronic timer if available) to reduce reaction time error.
- Ensure the track is level and free from obstacles so motion is uniform and not affected by bumps or friction changes.
- Release the car in a consistent way (no pushing) to avoid varying initial speeds.
- Measure distance accurately with a fixed metre scale and mark clear start/end lines.
- Repeat trials and discard obvious outliers before averaging.
Emphasize consistency and multiple trials to improve accuracy.
