JEE MAIN - Physics (2003 - No. 56)
The co-ordinates of a moving particle at any time 't' are given by x = $$\alpha $$t3 and y = βt3. The speed to the particle at time 't' is given by
$$3t\sqrt {{\alpha ^2} + {\beta ^2}} $$
$$3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$
$${t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$
$$\sqrt {{\alpha ^2} + {\beta ^2}} $$
Explanation
Given that $$x = \alpha {t^3}\,\,\,\,$$ and $$\,\,\,\,y = \beta {t^3}$$
$$\therefore$$ $${v_x} = {{dx} \over {dt}} = 3\alpha {t^2}\,\,\,\,$$
and$$\,\,\,\,\,{v_y} = {{dy} \over {dt}} = 3\beta {t^2}$$
$$\therefore$$ $$v = \sqrt {v_x^2 + v_y^2} $$
$$ = \sqrt {9{\alpha ^2}{t^4} + 9{\beta ^2}{t^4}} $$
$$ = 3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$
$$\therefore$$ $${v_x} = {{dx} \over {dt}} = 3\alpha {t^2}\,\,\,\,$$
and$$\,\,\,\,\,{v_y} = {{dy} \over {dt}} = 3\beta {t^2}$$
$$\therefore$$ $$v = \sqrt {v_x^2 + v_y^2} $$
$$ = \sqrt {9{\alpha ^2}{t^4} + 9{\beta ^2}{t^4}} $$
$$ = 3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$
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