Question

The coordinates of a moving particle at time $t$ are given by $x=c{t}^2$ and $y=b{t}^2$. The speed of the particle is given by

• A. $2t(c+b)$
• B. $2t\sqrt{(c^2-b^2)}$
• C. $t\sqrt{(c^2+b^2)}$
• D. $2t\sqrt{(c^2+b^2)}$

Answer 1

The correct option is D $2t\sqrt{(c^2+b^2)}$

$\frac{dx}{dt}=2ct$ and $\frac{dy}{dt}=2bt$
$\therefore v=\sqrt{\{\left(\frac{d{x}^2}{dt}\right)+{\left(\frac{dy}{dt}\right)}^2\}}$
$=\sqrt{\{{\left(2ct\right)}^2+{\left(2bt\right)}^2\}}$
$=2t\sqrt{(c^2+b^2)}$