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# A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10 m/s with respect to the water, in a direction perpendicular to the river. (a) Find the time taken by the boat to reach the opposite bank. (b) How far from the point directly opposite to the starting point does the boat reach the opposite bank?

6 years ago
Sol. a) Here the boat moves with the resultant velocity R. But the vertical component 10 m/s takes him to the opposite shore. Tan θ = 2/10 = 1/5 Velocity = 10 m/s distance = 400 m Time = 400/10 = 40 sec. b) The boat will reach at point C. In ∆ABC, tan θ = BC/AB = BC/400 = 1/5 ⇒ BC = 400/5 = 80 m.
Parikshit
13 Points
4 years ago
Sol. a) Here the boat moves with the resultant velocity R. But the vertical component 10 m/s takes him to the opposite shore. Tan θ = 2/10 = 1/5 Velocity = 10 m/s distance = 400 m Time = 400/10 = 40 sec. b) The boat will reach at point C. In ∆ABC, tan θ = BC/AB = BC/400 = 1/5 ⇒ BC = 400/5 = 80 m.
Maya
13 Points
3 years ago
Time =displacement (in y -direction) ————————————— Speed(in y direction)Time=. 400 ,—— 10 =40 sNow,Distance(in x direction) =s×t =2m/s×40s=80m