Question

A particle moves in the xy-plane with constant acceleration a directed along the negative y-axis. The equation of path of the particle has the form $y=bx-c{x}^2,$ where $b$ and $c$ are positive constants. The velocity $v$ of the particle at the origin of coordinates will be

• A. $\sqrt{\frac{a}{2c}(1+b^2)}$
• B. $\sqrt{\frac{a}{4c}(1+b^2)}$
• C. $\sqrt{\frac{3a}{2c}(1-b^2)}$
• D. none of these

Answer 1

Answer: (A)

$\sqrt{\frac{a}{2c}(1+b^2)}$

$x=v_{x}t$

$v_y=\frac{dy}{dt}=b\frac{dx}{dt}-2cx\frac{dx}{dt}=b{v}_x-2cx{v}_x$

therefore $v_y$ at x =0 y =0 is

$v_y=b{v}_x$

$S=ut+\frac{1}{2}a{t}^2$

$y=v_{y}t-0.5a{t}^2$

$v_y=b{v}_x$

$0.5a=c{v}_x^2$

$v=\sqrt{v_x^2+v_y^2}=\sqrt{v_x^2+(b{v}_x{)}^2}=\sqrt{v_x^2(1+b^2)}=\sqrt{\frac{a}{2c}(1+b^2)}$