1.

A man walks on a straight road from his home to a moardet `3 km` away with a speed of`6 km//h`. Finding the market closed, he instantly turns and walks back with a speed of `9 km//h`. What is the (a) magnitude of average velocity and (b) averge speed of the man, over the interval of time (i) `0 to 30 min`, (ii) `0to 50 min`, (iii) `0 to 40 min`?

Answer» Time taken by man ti go fro m his home to market,
`t_(1) =(distance )/(speed) =(3 km )/(6 km //h) =1/2 h =30 min`
Time taken by man to go from market to home,
`t_(2) =(3 km)/(9 km//h) =1/3 h =20 min`
Total time taken`=t_(1) +t_(2) =1/2 +1/3 =-5/6 h`
`=50 min`
(i) `0 to 30 min`
(a) Average velocity `=(displacement )/(time)`
`=(3.0 km)/((1//2) h) =6 km //h`
(b) Average speed =(path lenght)/(time) =(3.0 km )/((1//2) h)`
`=6 km //h`
(ii) `0 to 50 min`
Total path lenth coverad `=3.0 +3.0`
`=6.0 km`
Total displacement `=(displacement 0 /(time)`
Total displacement `=zero
(a) Average velocity `=(displacrmrnt)/(time)`
`=0/((5//6)h) =0`
(b) Average speed `=(total path length)/(time)`
`=(6 km )/((5//6)j) =7.2 km //h`
(iii) `0 to 40 min`
Distance moved in `30 min (from home to market) =3.0 km`
Distance moved in `10 min (from market to home) with speed `9 km //h` `=9xx (10)/(60) =1.5 km` .
So, displacement `=3.0 -1.5 =1.5 km`
Total path length travelled `3.0+ 1.5`
`=4.5 km`
(a) Average velocity`=(1.5km)/((400//60)h) =2.25 km //k` (b) Average speed `=(4.5 km)/9(40.60)h) =6.75 km //h`.


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