A man crosses a river in a boat. If he cross the river in minimum time he takes 10 minutes with a drift 120 m. if he crosses the river taking shortest path. He takes 12.5 minute find the velocity of boat with respect to water
Answers
Let the width of the river be = d
The minimum time taken to cross the river = t₁
The minimum time taken to cross the shortest distant = t₂
The horizontal drift for the first case = x
In our case :
t₁ = 10
t₂ = 12.5
x = 120 m
The formula to use is :
d = xt₂ / √(t₂ - t₁) (t₁ + t₂)
d = (120 × 12.5) / √(12.5 - 10)(12.5 + 10)
d = 1500 / √56.25
d = 1500/7.5 = 200
= 200m
Answer:
Explanation:
Width = d
Man = V
River = v
In minimum time, he goes perpendicular to the river.
d/V = 10 min------(1)
v x 10 min = 120 m, so v = 12 m /min
In minimum distance, the resultant of the velocities is perpendicular to river
If he is at angle x with the perpendicular
V sinx = v = 12-----(2)
V cosx = d / 12.5-----(3)
Squaring and adding (2) and (3)
2V^2 = 144 + d^2 / 156.25 -----(4)
Putting value of V from (1) in (4),
d^2 /50 = 144 + d^2 /156.25
d^2 = 10500 m
d = 100m
There could be some calculation error, but the method is correct.
Explanation: