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 - cx^2, where b and c are positive constants. Find the velocity of the particle at the origin of coordinates.

Answer 1

Therefore the velocity of the particle at the origin is $\sqrt{\frac{a}{2c}(1+b^2)}$

Explanation:

Given equation of path of the particle is

y = bx -cx²

Differentiating with respect to t

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

Let  $\frac{dy}{dt}=v_y$   and  $\frac{dx}{dt}=v_x$

$v_y=bv_x-2cxv_x$....(1)

At origin x=0 ,y=0

$v_y=bv_x-2c.0.v_x$

$\Rightarrow v_y=bv_x$....(2)

The acceleration of the particle is -a along y axis.

It means $\frac{dv_y}{dt}=-a$ and $\frac{dv_x}{dt}= 0$ [ since the acceleration along y axis]

Using (1) to find the acceleration,

∴$v_y=bv_x-2cxv_x$

Differentiating with respect to t

$\frac{dv_y}{dt}=b \frac{dv_x}{dt}-2c v^2_x-2cx\frac{dv_x}{dt}$

$\Rightarrow -a=b.0-2cv^2_x-2cx.0$    [since $\frac{dv_y}{dt}=-a$  and $\frac{dv_x}{dt}= 0$]

$\Rightarrow v_x=\sqrt{\frac{a}{2c}$....(3)

The velocity of the particle at origin is

$v=\sqrt{v^2_x+v^2_y$

Putting $v_y=bv_x$  [ from (2)]

$v=\sqrt{v^2_x+(bv_x)^2}$

  $=\sqrt{v^2_x(1+b^2)}$

  $=\sqrt{\frac{a}{2c}(1+b^2)}$     [putting $v_x=\sqrt{\frac{a}{2c}$  ]

Therefore the velocity of the particle at the origin is $\sqrt{\frac{a}{2c}(1+b^2)}$.