3.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h^–1 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^–1. Choose suitable scales and plot the xt graph of her motion.

<p><strong>3.3 </strong>Office distance = 2.5 km</p><p>Walking speed = 5 kph</p><p>Time taken to reach office = 5/2.5 = 0.5 h</p><p>OA is the path to reach office</p><p>AB is the duration of office stay</p><p>Auto speed = 25 kph</p><p>Time taken by auto = 2.5/25 h = 6 minutes</p><p>BC is the path for returning</p><p dir="ltr">On the position-time graph, this is a straight line sloping downwards. It starts at the point representing 5:00 PM and 2.5 km, and ends at 5:06 PM at 0 km. It tells us that the woman has arrived home.&nbsp;</p><p dir="ltr">The slope of this line represents&nbsp;&lt;a href="https://www.shiksha.com/preparation/physics-motion-in-straight-line-instantaneous-velocity-and-speed-5436-tp"&gt;instantaneous velocity&lt;/a&gt;. Because the return trip is by the auto at a much higher speed (25 km/h), the slope of line BC is significantly steeper than the line for the outward journey. A steeper slope on a position-time graph denotes a higher speed.</p><p dir="ltr"><strong><picture><img src="https://images.shiksha.com/mediadata/images/articles/1728556604phpih6mUv.jpeg" alt="" width="352" height="164"></picture></strong></p><p dir="ltr"><br>Master such questions with our &lt;a href="https://www.shiksha.com/preparation/physics-ncert-solutions-class-11th-motion-in-a-straight-line-4309-ns"&gt;NCERT Solutions for Motion in a Straight Line&lt;/a&gt;.&nbsp;</p>

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