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 ?

Explanation

Time taken be man to go from his home to market, `" " t_(1)=("distance")/("speed")=2.5/5=1/2 h`
Time taken be man to go from market to his home, `" " t_(2)=2.5/7.5=1/3h`
`:.` Total time taken `=t_(1)+t_(2)=1/2+1/3=5/6 h=50` min
(i) 0 to 30 min
Average velocity `=("displacement")/("time interval")=2.5/(30/60)=5km//h` towards market
Average speed `=("distance")/("time interval")=2.5/(30/60)=5 km//h`
(ii) 0 to 50 min
Total displacement =zero so average velocity `=0`
So, average speed `=5/(50//60)=6 km//h`
Total distance travelled `=2.5+2.5=5 km`.
(iii) 0 to 40 min
distance covered in 30 min (from home to market)`=2.5 km`.
Distance covered in 10 min (from market to home) with speed `7.5 km//h`
`=7.5xx10/60=1.25 km`
So, displacement `=2.5-1.25=1.25 km` (towards market)
Distance travelled `=2.5+1.25=3.75 km`
Average velocity `=1.25/(40/60)=1.875 km//h` (towards market)
Average speed `=3.75/(40/60)=5.625 km//h`
Note : Moving body with uniform speed may have variable velocity. e.g. in uniform circular motion speed is constant but velocity is non-uniform.