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?

Explanation

Time taken by the man to reach the market from home, `t_1=2.5/5=1/2 h`=30 min
Time taken by the man to reach the home from market , `t_2=2.5/7.5=1/3 h`=20 min
Total time taken in the whole journey =30+20 = 50 min
Average velocity=`("Displacement")/("Time")=2.5/1` =5 km/h ....(a(i))
Average speed = `("Distance")/("Time")=(2.5)/(1/2)`=5 km/h ...(b(i))
Time = 50 min = `5/6` h
Net displacement = 0
Total distance = 2.5 + 2.5 = 5 km
Average velocity =`("Displacement")/("Time")`=0 (a(ii))
Average speed =`("Distance")/("Time")=5/(5/6)=`6km /h ...(b(ii))
Speed of the man = 7.5 km
Distance travelled in first 30 min = 2.5 km
Distance travelled by the man (from market to home) in the next 10 min
`=7.5xx10/60=1.25 km`
Net displacement = 2.5 – 1.25 = 1.25 km
Total distance travelled = 2.5 + 1.25 = 3.75 km
Average velocity =`1.25/((40/60))=(1.25xx3)/2`=1.875 km/h ....(a(iii))
Average speed =`3.75/((40/60))`=5.625 km/h ...(b(iii)