Question 96: Motion in 1D

Question

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

Accepted Answer

Given that $$x = \alpha {t^3}\,\,\,\,$$ and $$\,\,\,\,y = \beta {t^3}$$
$$	herefore$$ $${v_x} = {{dx} \over {dt}} = 3\alpha {t^2}\,\,\,\,$$
and$$\,\,\,\,\,{v_y} = {{dy} \over {dt}} = 3\beta {t^2}$$
$$	herefore$$ $$v = \sqrt {v_x^2 + v_y^2} $$
$$ = \sqrt {9{\alpha ^2}{t^4} + 9{\beta ^2}{t^4}} $$
$$ = 3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$

Answer

$$3t\sqrt {{\alpha ^2} + {\beta ^2}} $$

Answer

$$3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$

Answer

$${t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $$

Answer

$$\sqrt {{\alpha ^2} + {\beta ^2}} $$

Motion in 1D

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

Choose the correct answer: