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A train first travels for 30 min with a velocity 30 km h^(-1) and then for 40 min with a velocity 40 km h^(-1) in the same direction, Calculate : (i) the total distance travelled, (ii) the average velocity of the train. |
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Answer» Solution :Given `t_(1) =30` min `=(1)/(2) h ,v_(1) = 30 "Km h"^(-1)` `ta_(2)` =40 min `= (2)/(3) h ,v_(2)`= 40 km `h^(-1)` . (i) DISTANCE travelled = velocity `xx` time `:. S_(1) = v_(1) xxt_(1) = 30 xx(1)/(2) = 15 ` km `S_(2) = v_(2) xxt_(2) = 40 xx(2)/(3) = (80)/(3)` km Total distance travelled `S= S_(1) + S_(2)` `= 15 + (80)/(3) = (125)/(3) `km = 41.67 km (ii) Total time of JOURNEY `t=t_(1) +t_(2)` `= (1)/(2) +(2)/(3) = (7)/(6) h` `:.` Average velocity = `("Total distance travelled S")/("Total time of journey t")` `= (41.67 km)/((7//6)h)= 35.71 km h^(-1).` |
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