The position vector of point A changes with time as vec r(t)…

The position vector of point A changes with time as $\vec r(t) = bt\,\vec i - ct^2\,\vec j$, where $b$ and $c$ are positive constants. Find: (a) the equation of the trajectory $y = y(x)$; (b) the velocity and acceleration vectors and their magnitudes as functions of time.

Explanation of Method

Trajectory, velocity and acceleration of a point.

Step 1: (a) Trajectory

The coordinates are $x = bt$ and $y = -ct^2$. Eliminate the time: $t = x/b$, so

\[ y = -c\left(\frac{x}{b}\right)^2. \]

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

This is a parabola opening downward with its vertex at the origin (for $t \ge 0$, only its right half $x \ge 0$ is travelled).

Step 2: (b) Velocity

\[ \vec v = \frac{d\vec r}{dt} = b\,\vec i - 2ct\,\vec j, \]

\[ v = \sqrt{b^2 + 4c^2t^2}. \]

Step 3: (b) Acceleration

\[ \vec a = \frac{d\vec v}{dt} = -2c\,\vec j, \]

\[ a = 2c. \]

\[\begin{gathered} \vec v = b\,\vec i - 2ct\,\vec j, \quad v = \sqrt{b^2 + 4c^2t^2} \\ \vec a = -2c\,\vec j, \quad a = 2c \end{gathered}\]

$x$ $y$ $O$ $b$ $2ct$ $\vec a$ $y = -\dfrac{c}{b^2}x^2$

v_x = b (constant)
v_y = −2ct (grows)
velocity
acceleration −2c j

The point moves along a downward-opening parabola (below the $x$-axis, where $y < 0$), like a ball thrown horizontally. The horizontal velocity stays $b$; the downward velocity grows in proportion to time; the acceleration is the same everywhere and points in the $-y$ direction.

Result

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

\[\boxed{\,\begin{gathered} \vec v = b\,\vec i - 2ct\,\vec j, \quad v = \sqrt{b^2 + 4c^2t^2} \\ \vec a = -2c\,\vec j, \quad a = 2c \end{gathered}\,}\]

Remark

Note. This is exactly the motion of a ball thrown horizontally with speed $b$ under gravity $g = 2c$: constant horizontal velocity, constant downward acceleration, and a parabolic path. The speed grows with time, but the acceleration never changes.