A man walks on a straight road from his home to market 2.5 km away with speed of 5 km/h Finding the market closed, he instantly turns and walks back with a speed of 7.5 km/h. What is the magnitude of average velocity and average speed of the man over the interval of time (i) 0 to 30 min (ii) 0 to 50 min (iii) 0 to 40 min ?

Options

A. `5 kmh^(-1)`
B. `(25)/(4)kmh^(-1)`
C. `(30)/(4)kmh^(-1)`
D. `(45)/(8)kmh^(-1)`

Answer

C

Explanation

Man walks from his home to market with a speed of `5 kmh^(-1)`. Distance `= 2.5 km` and time `=(d)/(upsilon)=(2.5)/(5)=(1)/(2)` and he retuens back with speed of `7.5 kmh^(-1)` in rest a time of 20 min.
Distance `=7.5xx(20)/(60)=2.5 km`
So, Average speed `=(Total distance")/("Total time")=((2.5+2.5))/((50//60)h)=(30)/(4)km//h`