Question

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?

Answer 1

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.