First slide

Relative motion in 2 dimension

Question

River stream velocity grows in proportion to the distance from the bank and reaches its maximum velocity 2 ms^-1in the middle. Near the bank velocity is zero. The velocity of a swimmer in still water is 5 ms^-1 and is directed perpendicular to river stream. The width of river is 100 m. The drifting in swimmer is 5n meter. The value of n is ______.

Difficult

Solution

$v_{x} = k y and v_{y} = 5$
$∴ y = 5 t, At middle point, 50 = 5 t$
$\begin{array}{l} ∴ t = 10 s \\ ∵ v_{x} = k y \end{array}$
$At middle, v_{x} = 2 ∴ k = \frac{2}{50}$
$\begin{array}{l} ∴ v_{x} = \frac{2}{50} y = \frac{y}{25} = \frac{5 t}{25} = \frac{t}{5} \\ \Rightarrow \frac{d x}{d t} = \frac{t}{5} \\ \Rightarrow \int_{0}^{x_{0}} d x = \int_{0}^{10} \frac{t}{5} d t \end{array}$
$∴ x_{0} = 10 m up to middle$
$∴ Total drifting = 2 \times 10 = 20 m$
$∴ 5 n = 20, n = 4$

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