A woman starts from her home at 9.00 am, walks with a speed of 5 km h on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h. Choose suitable scales and plot the graph of her motion.
Step 1: Calculate time to reach office Time taken = Distance/Speed = 2.5 km / 5 km h⁻¹ = 0.5 h = 30 min She reaches office at 9:30 am.
Step 2: Duration at office From 9:30 am to 5:00 pm, she stays at office. Duration = 7.5 hours (stationary position at x = 2.5 km).
Step 3: Calculate return time Time taken = Distance/Speed = 2.5 km / 25 km h⁻¹ = 0.1 h = 6 min She returns home at 5:06 pm.
Step 4: Plot the x-t graph Suitable scales: Time axis (x-axis): 1 major division = 1 hour. Position axis (y-axis): 1 major division = 0.5 km.
The graph has three parts:
- OA: Straight line from (9:00 am, 0) to (9:30 am, 2.5 km) with slope = 5 km/h (walking)
- AB: Horizontal line from (9:30 am, 2.5 km) to (5:00 pm, 2.5 km) — zero velocity (stationary)
- BC: Steep straight line from (5:00 pm, 2.5 km) to (5:06 pm, 0) with slope = -25 km/h (return by auto)
The return slope is steeper than forward journey slope, indicating higher speed.
Explanation
This question from NCERT Physics Chapter 2 (Motion in a Straight Line) tests understanding of position-time graphs. The student must calculate time intervals for each segment and plot accordingly. The x-t graph shows position on y-axis and time on x-axis. A straight line with positive slope indicates motion away from origin, horizontal line indicates rest, and line with negative slope indicates return motion. The steepness of slope represents speed magnitude.
Solution Steps
Step 1: Calculate walking time = 2.5/5 = 0.5 h, reaches office at 9:30 am
Step 2: Stationary at office from 9:30 am to 5:00 pm (7.5 hours)
Step 3: Calculate return time = 2.5/25 = 0.1 h = 6 min, reaches home at 5:06 pm
Step 4: Plot three segments: OA (walking), AB (stationary), BC (return by auto)