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Two pesons P and Q crosses the river starting from point A on side to excatly opposite point B on the other bank of the river. The person P crosses the river in the shortest path. The persons Q crosses the river in shortest time and walks back to point B. Velocity of river is 3 kmph and speed of each boat is 5kmph w.r.t river. If the two persons reach the point B in the same time, then the speed of walk of Q is |
Answer» Solution : `t_(p) = (d)/(sqrt(V_(B)^(2)-V_(W)^(2)))=(d)/(sqrt(5^(2)-3^(2)))=(d)/(4)` `t_(Q)=(d)/(V_(B))=(d)/(5), t_(P)=t_(Q)+Deltat` `(d)/(4)=(d)/(5)+(X)/(V_(man)), "" But x=V_(W)(d)/(V_(B))` `(d)/(4)=(d)/(5)+(V_(W)d)/(V_(B)V_(man)),""(canceld)/(4)=(canceld)/(5)+(3canceld)/((5)V_(man))` `(1)/(4)-(1)/(5)=(3)/(5V_(man)),""(1)/(20)=(3)/(5V_(man))` `V_(man) = ((3)(20))/(5)=12kmph` |
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