Question

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

• A. $2t(c+b{)}^{0.5}$
• B. $2t(c^2-b^2{)}^{0.5}$
• C. $t(c^2-b^2{)}^{0.5}$
• D. $2t(c^2+b^2{)}^{0.5}$

Answer 1

The correct option is D $2t(c^2+b^2{)}^{0.5}$

$x=c{t}^2$ & $y=b{t}^2$

So,

velocity in x direction, $v_x=\frac{dx}{dt}=2ct$

Velocity in y direction, $v_y=\frac{dy}{dt}=2bt$

So, net velocity

$v=\sqrt{v_x^2+v_y^2}$

$=\sqrt{4c^2{t}^2+4b^2{t}^2}$

$=2t\sqrt{c^2+b^2}$