A body moves in the xy -plane with velocity vec v = b vec i…

A body moves in the $xy$-plane with velocity $\vec v = b\,\vec i + cx\,\vec j$, where $b$ and $c$ are positive constants. At the start it was at the origin. Find the equation of the trajectory $y = y(x)$.

Explanation of Method

Trajectory from a position-dependent velocity.

Step 1: The $x$-motion

The $x$-component of velocity is constant, so starting from $x = 0$: $\frac{dx}{dt} = b \;\Longrightarrow\; x = bt$.

Step 2: The $y$-motion

Substitute $x = bt$ into the $y$-component: $\frac{dy}{dt} = cx = cb\,t$.

This is uniform acceleration $cb$ from rest ($y = 0$ at $t = 0$), so

\[ y = \tfrac12\,cb\,t^2. \]

Step 3: Eliminate time

With $t = x/b$: $y = \tfrac12\,cb\left(\frac{x}{b}\right)^2$.

\[y = \frac{c}{2b}\,x^2\]

Check

The slope of the path must equal the ratio of the velocity components: $\dfrac{dy}{dx} = \dfrac{c}{b}\,x$ and $\dfrac{v_y}{v_x} = \dfrac{cx}{b}\ \checkmark$. The path passes through the origin $\checkmark$

$y$ $x$ $O$ $b$ $c$ $b$ $2c$ $y = \dfrac{c}{2b}x^2$ $\vec v = b\,\vec i + cx\,\vec j$ $\text{(drawn for } b = c = 1)$

v_x = b
v_y = cx
velocity (tangent to the path)

The horizontal velocity is always $b$, while the vertical velocity grows with $x$: at $x = 2$ it is twice as large as at $x = 1$. The velocity is always tangent to the path, so the path gets steeper and steeper: a parabola starting at the origin.

Result

\[\boxed{\,y = \frac{c}{2b}\,x^2\,}\]