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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?`
5 years ago 396 Points
```							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.

```
5 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.
```
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
```
2 years ago
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