1.14 Velocity  (Page 6/6)

Solution : The velocity of B is not an independent velocity. It is tied to the velocity of the particle “A” as two particles are connected through a rigid rod. The relationship between two velocities is governed by the inter-particles separation, which is equal to the length of rod.

The length of the rod, in turn, is linked to the positions of particles “A” and “B” . From figure,

$$x=\sqrt{\left(L^2-y^2\right)}$$

Differentiatiting, with respect to time :

$$\Rightarrow v_x=\frac{dx}{dt}=-\frac{2y}{2\sqrt{\left(L^2-y^2\right)}}X\frac{dy}{dt}=-\frac{y{v}_y}{\sqrt{\left(L^2-y^2\right)}}=-v_y\tan \theta$$

Considering magnitude only,

$$\Rightarrow v_x=v_y\tan \theta =10\tan {60}^0=10\sqrt{3}\frac{m}{s}$$

Solution : In order to prove as required, we shall use the fact that scalar (dot) product of two perpendicular vectors is zero. Now, we need to find the expression of velocity to evaluate the dot product as intended. We can obtain the same by differentiating the expression of position vector with respect to time as :

$$\begin{array}{l}\mathbf{v}=\frac{d\mathbf{r}}{dt}=-a\omega \sin \omega t\mathbf{i}+a\omega \cos \omega t\mathbf{j}\end{array}$$

To check whether velocity is perpendicular to the position vector, we evalaute the scalar product of r and v , which should be equal to zero.

$$\begin{array}{l}\mathbf{r}\mathbf{.}\mathbf{v}=0\end{array}$$

In this case,

$$\begin{array}{l}\Rightarrow \mathbf{r}\mathbf{.}\mathbf{v}=(a\cos \omega t\mathbf{i}+a\sin \omega t\mathbf{j})\mathbf{.}(-a\omega \sin \omega t\mathbf{i}+a\omega \cos \omega t\mathbf{j})\\ \Rightarrow -a^2\omega \sin \omega t\cos \omega t+a^2\omega \sin \omega t\cos \omega t=0\end{array}$$

This means that the angle between position vector and velocity are at right angle to each other. Hence, velocity is perpendicular to position vector. It is pertinent to mention here that this result can also be inferred from the plot of motion. An inspection of position vector reveals that it represents uniform circular motion as shown in the figure here.

The position vector is always directed radially, whereas velocity vector is always tangential to the circular path. These two vectors are, therefore, perpendicular to each other.

Solution : We draw the figure to illustrate the situation. Here, car travels the linear distance (AB + CD) along the direction in which it moves, by which time the pedestrian travels the linear distance BD. Let pedestrian travels at a speed “v” along BD, which makes an angle “θ” with the direction of car.

We must understand here that there may be number of combination of angle and speed for which pedestrian will be able to safely cross before car reaches. However, we are required to find the minimum speed. This speed should, then, correspond to a particular value of θ.

We can also observe that pedestrian should move obliquely. In doing so he/she gains extra time to cross the road.

From triangle BCD,

$$\begin{array}{l}\tan (90-\theta )=\cot \theta =\frac{\mathrm{CD}}{\mathrm{BC}}=\frac{\mathrm{CD}}{2}\\ \Rightarrow \mathrm{CD}=2\cot \theta \end{array}$$

Also,

$$\begin{array}{l}\cos (90-\theta )=\sin \theta =\frac{\mathrm{BC}}{\mathrm{BD}}=\frac{2}{\mathrm{BD}}\\ \Rightarrow \mathrm{BD}=\frac{2}{\sin \theta }\end{array}$$

According to the condition given in the question, the time taken by car and pedestrian should be equal for the situation outlined above :

$$\begin{array}{l}t=\frac{4+2\cot \theta }{8}=\frac{\frac{2}{\sin \theta }}{v}\end{array}$$

$$\begin{array}{l}v=\frac{8}{2\sin \theta +\cos \theta }\end{array}$$

For minimum value of speed, $\frac{dv}{d\theta }=0$ ,

$$\begin{array}{l}\Rightarrow \frac{dv}{d\theta }=\frac{-8x(2\cos \theta -\sin \theta )}{{(2\sin \theta +\cos \theta )}^2}=0\\ \Rightarrow (2\cos \theta -\sin \theta )=0\\ \Rightarrow \tan \theta =2\end{array}$$

In order to evaluate the expression of velocity with trigonometric ratios, we take the help of right angle triangle as shown in the figure, which is consistent with the above result.

From the triangle, defining angle “θ”, we have :

$$\begin{array}{l}\sin \theta =2\surd 5\end{array}$$

and

$$\begin{array}{l}\cos \theta =\frac{1}{\surd 5}\end{array}$$

The minimum velocity is :

$$\begin{array}{l}v=\frac{8}{2x2\surd 5+\frac{1}{\surd 5}}=\frac{8}{\surd 5}=3.57m/s\end{array}$$