A man walks on a straight road from his home to a market ` 2.5 km` away with a 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 (a) magnitude of average velocity and (b) 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. 0,0
B. `6 km h^(-1), 0`
C. 0, `6 km^(-1)`
D. `6 km h^(-1), 6 km h^(-1)`

Answer

B

Explanation

Time taken by the boy to go from his home to the market, `t_1 = (2.5 ("km"))/(5 "km" h^(-1)) = 1/2` h
Time taken by the boy to return back from the market to his home, `t_2= (2.5 km)/(7.5 km h^(-1))= 1/3` h
`therefore` Total time taken = `t_1 + t_2 = 1/2 h + 1/3 h = 5/6 h = 50` min
In t = 0 to 50 min.
Total distance travelled = 2.5 km + 2.5 km = 5 km)
Displacement = 0 (As the boy returns back home)
`therefore` Average speed `= ("Distance travelled") /("Time taken")`= `(5 "km")/(5/6h) = 6 km h^(-1)`
Average velocity `= ("Displacement")/("Time taken") = 0`