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A man walks on a straight road from his home to a market 2.5 km away with a speed of 5kmh-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5kmh-1. What is the magnitude of average velocity, and average speed of the man over the following intervals of time:1. 0 to 30 min,2. 0 to 50 min,3. 0 to 40 min? |
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Answer» Average speed over the interval of time from 0 to 30min = \(\frac{20km}{30min} = \frac{2.5 km}{\frac{1}{2}h}\) = 5 kmh-1 Magnitude of average velocity over the interval of time from 0 to 30 min is 5kmh-1. This is because the “distance travelled” and the ‘magnitude of displacement’ over the interval of time from 0 to 30 min are equal. Distance covered from 30 to 50 minutes = 7.5kmh-1 × \(\frac{20}{60}\) h = 2.5km Total distance covered from 0 to 50 minute = 2.5 km+ 2.5 km = 5km Total time = 50min = \(\frac{50}{60}\) h = \(\frac{5}{6}\) h Average speed over the interval of time from 0 to 50min = \(\frac{5km}{5/6h}\) = 6kmh-1 The displacement over the interval of time from 0 to 50 min is zero. So the magnitude of average velocity is zero. Distance covered from 30 to 40 min = 7.5kmh-1 × \(\frac{1}{6}\) h = 1.25km Total distance covered from 0 to 40 minute = 2.5 km + 1.25 km = 3.75km Average speed over the interval of time from 0 to 40min = \(\cfrac{3.75km}{\frac{40}{60}h}\) = 5.625kmh-1 The “magnitude of displacement” is (2.5 -1.25) km, i.e., 1.25 km. Time interval = \(\frac{2}{3}\)h The ‘magnitude of average velocity’ = \(\frac{1.25 km}{2/3h}\), = 1.875 kmh-1. |
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