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InterviewSolution
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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 ? |
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Answer» Time tanen by man to go from his home to market, ` t _1 = (distance )/(speed) = (2.5)/ 5 = 1/2 h` Time taken by man to go from market to his home, ` T_2 = (2.5) /( 7. 5) = 1/3 h` :. Total time taken ` t_1 + t_2 + 1/2 = 5/6 = 5/6 h = 50 min. (i) to 30 min` (a) Acerage velocity = (dispacement )/(time ) = ( 2.5)/ (1//2) = 5 km// h (b) Avereage speed = (distance ) /(time ) = ( 2.5) /(1//2) = 5 km //h` (ii) to ` 50 min` Time distance travelled` = 2.5 + 2. 5 = 5 km ` Total displacment = zero (a) Average velocity = (displacemt )/( time) =0 (b) Average speed = (distace) /(5//6) = 6 km //h` ( iii) ` 0 to 40 min` . Distance moved in ` 10 min` (from ome to market) = 2.5 km`. Distance moved in ` 10 min (from market to home ) with speed ` 7.5 km//h = 7. 5 xx (10)/ (60) = 1. 2 5 km` so displacemnt` = 2.5 - 1. 25 = 1. 25 km` Distanc travelled `= 2. 5 + 1. 25 = 3. 75 km ` ltbRgt (a) Average velocity `= ( 1.250/ (( 40 //60)) = 1.875 km //h` ltbRgt (b) Average speed ` = (3. 75)/ ( ( 40 // 60 )) =5. 625 km //h`. |
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