The position vector of a particle is vec r(t) = (A + Bt^2)…

The position vector of a particle is $\vec r(t) = (A + Bt^2)\,\vec i + Ct\,\vec j$, where $A$, $B$, $C$ are constants. Find (a) the equation of the trajectory; (b) the velocity and acceleration vectors and their magnitudes; (c) the magnitudes of the tangential and normal components of the acceleration.

Explanation of Method

Tangential and normal acceleration on a parabolic path.

Step 1: (a) Trajectory

From $x = A + Bt^2$ and $y = Ct$, eliminate $t = y/C$: $x = A + \frac{B}{C^2}\,y^2$.

A parabola opening along the $x$-axis with its vertex at $(A, 0)$.

Step 2: (b) Velocity and acceleration

\[ \vec v = 2Bt\,\vec i + C\,\vec j, \]

\[ v = \sqrt{4B^2t^2 + C^2}, \]

\[ \vec a = 2B\,\vec i, \]

\[ a = 2|B| \ \text{(constant)}. \]

$y$ $x$ $A$ $\vec v$ $\vec a = 2B\,\vec i$ $a_t$ $a_n$ $x = A + \tfrac{B}{C^2}y^2$ $\text{(drawn for } A = 1,\ B = 1,$ $C = 2,\ t = 1\ \mathrm{s})$

velocity
acceleration
tangential part
normal part

The acceleration is constant and points along $+x$. Splitting it into a part along the velocity (tangential) and a part across it (normal) gives $a_t$ and $a_n$. In the drawn example both equal $\sqrt2\ \mathrm{m/s^2}$.

Step 3: (c) Tangential acceleration

\[ a_t = \frac{dv}{dt} = \frac{4B^2t}{\sqrt{4B^2t^2 + C^2}}. \]

Step 4: (c) Normal acceleration

\[ a_n = \sqrt{a^2 - a_t^2} = \sqrt{4B^2 - \frac{16B^4t^2}{4B^2t^2 + C^2}} = \sqrt{\frac{4B^2C^2}{4B^2t^2 + C^2}}. \]

\[a_t = \frac{4B^2t}{\sqrt{4B^2t^2 + C^2}}, \qquad a_n = \frac{2|BC|}{\sqrt{4B^2t^2 + C^2}}\]

Check

$ a_t^2 + a_n^2 = \dfrac{16B^4t^2 + 4B^2C^2}{4B^2t^2 + C^2} = 4B^2 = a^2\ \checkmark $

At $t = 0$ (the vertex) $a_t = 0$ and $a_n = 2|B|$: all of the acceleration is normal there $\checkmark$

Result

\[\boxed{\,x = A + \frac{B}{C^2}\,y^2\,}\]

\[\boxed{\,a_t = \frac{4B^2t}{\sqrt{4B^2t^2 + C^2}}, \qquad a_n = \frac{2|BC|}{\sqrt{4B^2t^2 + C^2}}\,}\]

Remark

Note. The original solution stopped at the unsimplified square root; bringing it to a common denominator gives the compact form above. (The statement's "$A + Bt^2\vec i$" means that $A$ is part of the $x$-coordinate.)