Question

If line with y=0 is moving with velocity vĴ and line y=x is moving with velocity (v/√2)*(i^ - Ĵ), then find speed of intersection point of two lines. ( here i^ means i cap and Ĵ means j cap)

Answer 1

If line with $y = 0$ is moving with velocity $v\bar{J}$ and line $y = x$ is moving with velocity $(v / v2)^{*}(i^{\wedge} - \bar{J})$ , then find speed of intersection point of two lines. (here $i^{\wedge}$ means i cap and $\bar{J}$ means j cap)

Equation of the first line $(y = y_0 + vt, y_0 = 0)$ :

$$y - v t = 0$$

Equation of the second line $(y = y_0 - \frac{v}{\sqrt{2}} t, x = x_0 + \frac{v}{\sqrt{2}} t, x_0 = y_0)$ :

$$y + \frac {v}{\sqrt {2}} t = x - \frac {v}{\sqrt {2}} t$$

The system of equations allowing to find intersection point:

$$\left\{ \begin{array}{c} y - v t = 0 \\ y + \frac {v}{\sqrt {2}} t = x - \frac {v}{\sqrt {2}} t \end{array} \right.$$

or:

$$\left\{ \begin{array}{l} y = v t \\ x = y + \sqrt {2} v t \end{array} \right. \Rightarrow \left\{ \begin{array}{l} y = v t \\ x = (1 + \sqrt {2}) v t \end{array} \right.$$

Therefore, its speed:

$$V = \sqrt {\left(\frac {d y}{d t}\right) ^ {2} + \left(\frac {d y}{d t}\right) ^ {2}} = v \sqrt {\left(1 + \sqrt {2}\right) ^ {2} + 1 ^ {2}} = v \sqrt {4 + 2 \sqrt {2}}$$

Answer: $v\sqrt{4 + 2\sqrt{2}}$