Question

two cars are moving with same velocity c/3.these are perpendicular to each other .find relative velocity.

Answer 1

Let's denote the velocity of the first car $\mathbf{v}_1$ and velocity of the second car $\mathbf{v}_2$. The relativistic velocity-addition formula is the following (see https://en.wikipedia.org/wiki/Velocity-addition_formula#General_configuration):

$$\mathbf v_1 \oplus \mathbf v_2 \equiv \mathbf v =\frac{1}{1 - \frac{\mathbf v_1 \cdot \mathbf v_2}{c^2}}\left[\frac{\mathbf v_1}{\gamma_{v_2}} - \mathbf v_2 + \frac{1}{c^2}\frac{\gamma_{v_2}}{1+\gamma_{v_2}}(\mathbf v_1 \cdot \mathbf v_2)\mathbf v_2\right]\\$$

where $\gamma_{v_2} = \dfrac{1}{\sqrt{1 - \dfrac{v_2^2}{c^2}}} = \dfrac{1}{\sqrt{1 - \dfrac{c^2}{3^2\cdot c^2}}} = \dfrac{3}{\sqrt{8}}$.

As far as velocities are perpendicular to each other, $(\mathbf v_1 \cdot \mathbf v_2)\mathbf v_2 = 0$, then

$$\mathbf v =\frac{1}{1 - 0}\left[\frac{\mathbf v_1}{\gamma_{v_2}} - \mathbf v_2 \right] = \frac{\mathbf v_1}{\gamma_{v_2}} - \mathbf v_2 = \dfrac{\sqrt{8}}{3}\mathbf v_1-\mathbf v_2\\$$

The modulus of this relative velocity will be:

$$|\mathbf v|^2= \left( \dfrac{\sqrt{8}}{3}\mathbf v_1-\mathbf v_2 \right)\left( \dfrac{\sqrt{8}}{3}\mathbf v_1-\mathbf v_2 \right) = \dfrac{8}{9}|\mathbf v_1|^2 + |\mathbf v_2|^2 = \\ =\dfrac{8}{9}\dfrac{c^2}{9} + \dfrac{c^2}{9} = \dfrac{17c^2}{81}$$

$$v = \sqrt{\dfrac{17c^2}{81}} = \dfrac{\sqrt17}{9}c \approx 0.46c$$

Answer. 0.46c.