A man walks on a straight road from his home to a market 5.5 km away with a speed of 5.5 km `h^(-1)`. Finding the market closed he instantly turns and walks back with a speed of 7.5 km `h^(-1)`. What is the average speed and average velocity of the man between t = 0 to t = 104 min?

Options

A. 0,0
B. `6.36 km h^(-1) , 0`
C. `0, 6.36 km h^(-1)`
D. `6.36 km h^(-1) , 6.36 km h^(-1)`

Answer

B

Explanation

Time taken by the man to go from his home to the market , ` t_1 = (5.5 km)/(5.5 km h^(-1)) = 1.0 h`
Time taken by the man to return back from the market to his home , `t_2 =(5.5 km)/(7.5 km h^(-1)) = 0.73 h`
`therefore `Total Time taken `=t_1 + t_2 = 1.0 h + 0.73 h = 1.73 h = 104` min
In t = 0 to 104 min, total distance travelled = 5.5 km + 5.5 km = 11 km
Displacement = 0 (As the boy returns back home)
`therefore ` Average speed `=("Distance travelled")/("Time taken") = (11 km)/(1.73 h) = 6.36 km h^(-1)`
Average velocity `=("Displacement")/("Time taken") =0`