Question

A particle moves in the plane the plane xy with velocity V = $k_1$i + $k_2$ $\times$j, where and j are the unit vectors of the x and y axes, and $k_1$ and $k_2$ are constants. At the initial moment of time the particle was located at the point x = y = 0 then the equation of the particle's trajectory y (x) is -

• A. y = $\frac{k_1}{2k_2}$ $x^2$
• B. y = $\frac{k_2}{2k_1}$ $x^2$
• C. y = $\frac{2k_1}{k_2}$ $x^2$
• D. y = $\frac{2k_2}{k_1}$ $x^2$

Answer 1

Answer: (B)

y = $\frac{k_2}{2k_1}$ $x^2$

Velocity vector

$V=k_1\hat{i}+k_{2}x\hat{j}$

$\frac{dx}{dt}=k_1,\frac{dy}{dt}=k_{2}x\left(1\right)$

${\int }_0^{x}dx=k_1{\int }_0^{t}dt,x=k_{1}t\left(2\right)$

$dy=k_{2}xdt=k_2{k}_{1}dt$

${\int }_0^{y}dy=k_1{k}_2{\int }_0^{t}tdt,$

$y=\frac{1}{2}{k}_2{k}_1{t}^2$

$y=\frac{k_2}{2k_1}{x}^2$