3.6 Resultant motion (application)  (Page 3/2)

Problem : A person can swim at 1 m/s in still water. He swims to cross a river of width 200 m to a point exactly opposite to his/her initial position. If the water stream in river flows at 2 m/s in a linear direction, then find the time taken (in seconds) to reach the opposite point.

Solution : Let the direction of stream be x-direction and the direction across stream be y-direction. Let us also denote person with “A” and water stream with “B”.

To reach the point across, the person has to swim upstream at an angle such that the velocity of the person with respect to ground( ${\mathbf{v}}_A$ ) is across the direction of water stream. The situation is shown in the figure.

Here,

$$\begin{array}{l}\text{Speed of the person (A) with respect to stream (B) :}{v}_{AB}=1m\mathrm{/}s\\ \text{Speed of water stream (B) with respect to ground :}{v}_B=2m\mathrm{/}s\\ \text{Speed of the person (A) with respect to ground :}{v}_A=?\\ d=200m\end{array}$$

From the ΔOAB,

$$\begin{array}{l}{\mathrm{OQ}}^2={\mathrm{OP}}^2+{\mathrm{PQ}}^2\\ \Rightarrow \mathrm{OP}=\surd \{{\mathrm{OQ}}^2-{\mathrm{PQ}}^2\}\\ \Rightarrow t=\frac{d}{\surd \{{\mathrm{OB}}^2-{\mathrm{AB}}^2\}}\end{array}$$

It is clear from the denominator of the expression that for finite time, OB>AB. From the values as given in the question, OB<AB and the denominator becomes square root of negative number. The result is interpreted to mean that the physical event associated with the expression is not possible.The swimmer, therefore, can not reach the point, which is exactly opposite to his position. The speed of the swimmer should be greater than that of the stream to reach the point lying exactly opposite.

Note that we had explained the same situation in the module on the subject with the help of the value of "sinθ", which can not be greater than 1. We have taken a different approach here to illustrate the same limitation of the swimmer's ability to cross the water stream, showing that the interpretation is consistent and correct.

Problem : The direction of water stream in a river is along x – direction of the coordinate system attached to the ground. A swimmer swims across the river with a velocity ( $0.8\mathbf{i}+1.4\mathbf{j}$ ) m/s, as seen from the ground. If the river is 70 m wide, how long (in seconds) does he take to reach the river bank on the other side ?

Solution : We recognize here that the given velocity represents the resultant velocity ( ${\mathbf{v}}_A$ ) of the swimmer (A). The time to reach the river bank on the other side is a function of component velocity in y-direction.

Here,

$$\begin{array}{l}{v}_x=0.8m\mathrm{/}s\\ v_y=1.4m\mathrm{/}s\end{array}$$

$$\begin{array}{l}t=\frac{\mathrm{Width\; of\; the\; river}}{v_y}\\ t=\frac{70}{1.4}=50s\end{array}$$