3.3 Relative velocity in two dimensions (application)  (Page 2/2)

Problem : Rain drop appears to fall in vertical direction to a person, who is walking at a velocity 3 m/s in a given direction. When the person doubles his velocity in the same direction, the rain drop appears to come to make an angle 45° from the vertical. Find the speed of the rain drop.

Solution : This is a slightly tricky question. Readers may like to visualize the problem and solve on their own before going through the solution given here.

Let us draw the situations under two cases. Here, only the directions of relative velocities in two conditions are given. The figure on left represents initial situation. Here, the vector OP represents velocity of the person ( ${\mathbf{v}}_A$ ); OR represents relative velocity of rain drop with respect to person ( ${\mathbf{v}}_{\mathrm{BA}}$ ); OS represents velocity of rain drop.

The figure on right represents situation when person starts moving with double velocity. Here, the vector OT represents velocity of the person ( ${\mathbf{v}}_{\mathrm{A1}}$ ); OW represents relative velocity of rain drop with respect to person ( ${\mathbf{v}}_{\mathrm{BA1}}$ ). We should note that velocity of rain ( ${\mathbf{v}}_B$ ) drop remains same and as such, it is represented by OS represents as before.

According to question, we are required to know the speed of raindrop. It means that we need to know the angle “θ” and the side OS, which is the magnitude of velocity of raindrop. It is intuitive from the situation that it would help if consider the vector diagram and carry out geometric analysis to find these quantities. For this, we substitute the vector notations with known magnitudes as shown here.

We note here that

$$WR=UQ=4m∕s$$

Clearly, triangles ORS and ORW are congruent triangles as two sides and one enclosed angle are equal.

$$WR=RS=4m/s$$

$$OR=OR$$

$$\angle ORW=\angle ORS={90}^0$$

Hence,

$$\angle WOR=\angle SOR={45}^0$$

In triangle ORS,

$$\sin {45}^0=\frac{RS}{OS}$$

$$\Rightarrow OS=\frac{RS}{\sin {45}^0}=4\sqrt{2}m/s$$

Problem : Rain drop appears to fall in vertical direction to a person, who is wlaking at a velocity 3 m/s in a given direction. When the person doubles his velocity in the same direction, the rain drop appears to come to make an angle 45° from the vertical. Find the speed of the rain drop, using unit vectors.

Solution : It is the same question as earlier one, but is required to be solved using unit vectors. The solution of the problem in terms of unit vectors gives us an insight into the working of algebraic analysis and also let us appreciate the power and elegance of using unit vectors to find solution of the problem, which otherwise appears to be difficult.

Let the velocity of raindrop be :

$$v_B=ai+bj$$

where “a” and “b” are constants. Note here that we have considered vertically downward direction as positive.

According to question,

$$v_A=4i$$

The relative velocity of raindrop with respect to person is :

$$v_{BA}=v_B-v_A=ai+bj-4i=\left(a-4\right)i+bj$$

However, it is given that the raindrop appears to fall in vertical direction. It means that relative velocity has no component in x-direction. Hence,

$$a-4=0$$

$$a=4m/s$$

and

$$v_{BA}=bj$$

In the second case,

$$v_{A_1}=8i$$

$$v_{B{A}_1}=v_B-v_A=ai+bj-8i=\left(a-8\right)i+bj=\left(4-8\right)i+bj=-4i+bj$$

It is given that the raindrop appears to fall in the direction, making an angle 45° with the vertical.

$$\tan {135}^0=-1=-\frac{4}{b}$$

Note that tangent of the angle is measured from positive x – direction (90° + 45° = 135°) of the coordinate system.

$$b=4$$

Thus, velocity of the raindrop is :

$$v_B=ai+bj=-4i+bj$$

The speed of raindrop is :

$$\Rightarrow v_B=\sqrt{\left(4^2+4^2\right)}=4\sqrt{2}m/s$$