Question

A particle of mass m moves along a curve along a curve, $y=x^2$. When particle has x-coordinate as $\frac{1}{2}$ and x-component of velocity as 4 m/s then

• A. The position coordinate of particle are $(\frac{1}{2},\frac{1}{4})$
• B. The velocity of particle will be along the line 4x-4y-1=0
• C. The magnitude of velocity at that instant is $4\sqrt{2}m/s$

Answer 1

Answer: (A), (B), (C)

The magnitude of velocity at that instant is $4\sqrt{2}m/s$

(a)Given trajectory is $y=x^2$
$\therefore$ position when $x=\frac{1}{2}$,

$y=(\frac{1}{2}{)}^2=\frac{1}{4}m$
$\therefore position(\frac{1}{2},\frac{1}{4})$

(b) Slope at $(\frac{1}{2},\frac{1}{4})$
$\frac{dy}{dx}=2x\therefore atx=\frac{1}{2},\frac{dy}{dx}=1$
Velocity of the particle will be along the tangent at $(\frac{1}{2},\frac{1}{4})$

Equation of line, given slope and one point is:
$(y-y_1)=m(x-x_1)$
$\Rightarrow y-\frac{1}{4}=1(x-\frac{1}{2})$
$\Rightarrow \frac{4y-1}{4}=\frac{2x-1}{2}\Rightarrow 4y-1=4x-2$

$\Rightarrow 4y-4x+1=0$
velocity will be along the line, $4y-4x+1=0$

(c) Given $v_x=4m/sat,x=\frac{1}{2}$
Differentiating w.r.t. "t"
$y=x^2$
$\frac{dy}{dt}=2x\frac{dx}{dt}$
$\Rightarrow v_y=2x{v}_x$
$\Rightarrow atx=\frac{1}{2}$
$(v_y=\frac{2\times 1}{2}(4\left)\right)$
$\Rightarrow v_y=4m/s$
$\therefore$ magnitude of resultant velocity=$\sqrt{v_x^2+v_y^2}$
$=\sqrt{4^2+4^2}$
$=\sqrt{2\times 16}$
$=4\sqrt{2}$ m/s