3.2 Relative velocity in two dimensions  (Page 3/3)

Problem : A person is driving a car towards east at a speed of 80 km/hr. A train appears to move towards north with a velocity of 80√3 km/hr to the person driving the car. Find the speed of the train as measured with respect to earth.

Solution : Let us first identify the car and train as “A” and “B”. Here, we are provided with the speed of car (“A”) with respect to Earth i.e. " $v_A$ " and speed of train (“B”) with respect to “A” i.e $v_{\mathrm{BA}}$ .

$$\begin{array}{l}{v}_A=80\mathrm{km}\mathrm{/}\mathrm{hr}\\ v_{BA}=80\surd 3\mathrm{km}\mathrm{/}\mathrm{hr}\end{array}$$

We are required to find the speed of train (“B”) with respect to Earth i.e. ${\mathbf{v}}_B$ , . Fromequation of relative motion, we have :

$$\begin{array}{l}{\mathbf{v}}_{BA}={\mathbf{v}}_B-{\mathbf{v}}_A\\ \Rightarrow {\mathbf{v}}_B={\mathbf{v}}_{BA}+{\mathbf{v}}_A\end{array}$$

To evaluate the right hand side of the equation, we draw vectors “ ${\mathbf{v}}_{BA}$ ” and “ ${\mathbf{v}}_A$ ” and use parallelogram law to find the actual speed of the train.

$$\begin{array}{l}\Rightarrow v_B=\surd ({v_{BA}}^2+{v_A}^2)=\surd \{{(80\surd 3)}^2+{80}^2\}=160\mathrm{km}\mathrm{/}\mathrm{hr}\end{array}$$

Problem : A person, standing on the road, holds his umbrella to his back at an angle 30° with the vertical to protect himself from rain. He starts running at a speed of 10 m/s along a straight line. He finds that he now has to hold his umbrella vertically to protect himself from the rain. Find the speed of raindrops as measured with respect to (i) ground and (ii) the moving person.

Solution : Let us first examine the inputs available in this problem. To do this let us first identify different entities with symbols. Let A and B denote the person and the rain respectively. The initial condition of the person gives the information about the direction of rain with respect to ground - notably not the speed with which rain falls. It means that we know the direction of velocity ${\mathbf{v}}_B$ . The subsequent condition, when person starts moving, tells us the velocity of the person “A” with respect to ground i.e ${\mathbf{v}}_A$ . Also, it is given that the direction of relative velocity of rain “B” with respect to the moving person “A” is vertical i.e. we know the direction of relative velocity ${\mathbf{v}}_{BA}$ .

We draw three vectors involved in the problem as shown in the figure. OP represents ${\mathbf{v}}_A$ ; OQ represents ${\mathbf{v}}_B$ ; OR represents ${\mathbf{v}}_{BA}$ .

In ΔOCB,

$$\begin{array}{l}{v}_B=\mathrm{OR}=\frac{\mathrm{QR}}{\sin {30}^0}\\ \Rightarrow v_B=\frac{10}{\frac{1}{2}}=20m\mathrm{/}s\end{array}$$

and

$$\begin{array}{l}{v}_{BA}=\mathrm{OQ}=\frac{\mathrm{QR}}{\tan {30}^0}\\ \Rightarrow v_{BA}=\frac{10}{\frac{1}{\surd 3}}=10\surd 3m\mathrm{/}s\end{array}$$

Problem : Three particles A,B and C situated at the vertices of an equilateral triangle starts moving simultaneously at a constant speed “v” in the direction of adjacent particle, which falls ahead in the anti-clockwise direction. If “a” be the side of the triangle, then find the time when they meet.

Solution : Here, particle “A” follows “B”, “B” follows “C” and “C” follows “A”. The direction of motion of each particle keeps changing as motion of each particle is always directed towards other particle. The situation after a time “t” is shown in the figure with a possible outline of path followed by the particles before they meet.

This problem appears to be complex as the path of motion is difficult to be defined. But, it has a simple solution in component analysis. Let us consider the pair “A” and “B”. The initial component of velocities in the direction of line joining the initial position of the two particles is “v” and “vcosθ” as shown in the figure here :

The component velocities are directed towards eachother. Now, considering the linear (one dimensional) motion in the direction of AB, the relative velocity of “A” with respect to “B” is :

$$\begin{array}{l}{v}_{AB}=v_A-v_B\\ v_{AB}=v-(-v\cos \theta )=v+v\cos \theta \end{array}$$

In equilateral triangle, θ = 60°,

$$\begin{array}{l}{v}_{AB}=v+v\cos {60}^0=v+\frac{v}{2}=\frac{3v}{2}\end{array}$$

The time taken to cover the displacement “a” i..e. the side of the triangle,

$$\begin{array}{l}t=\frac{2a}{3v}\end{array}$$