Prompt: An object of mass $m$ travels along a parabola $y = x^2$ with constant speed $5$ units/sec. What is the force on the object due to acceleration at point $(\sqrt{2}, 2)$? Recall $\vec{F} = m\vec{a}$.
This was an "applied theory" problem I got incorrect on an generated exam I recently took. This question seems to be a similar idea, excepting that I need to work backwards from speed rather than velocity and that speed is constant. Here is what I understand:
- Since we're working in 2-D, $|\vec{v}(t)| = \sqrt{v_x^2 + v_y^2} = 5 \implies |\vec{v}(t)|^2 = 25 = v_x^2 + v_y^2$. I don't know if this helps me.
- The parabola can be reparameterized as a trajectory/displacement curve: $\vec{r}(t) = \langle t, t^2 \rangle$
- From the reparameterization, we can differentiate $\frac{d \vec{r}}{dt} = \vec{v}(t) = \langle 1, 2t \rangle$ and once more $\frac{d \vec{v}}{dt} = \vec{a}(t) = \langle 0, 2 \rangle$.
- If I compute the supposed magnitude of $\vec{v}$ at point $(\sqrt{2}, 2) \Longleftrightarrow t = \sqrt{2}$, I find $$ |\vec{v}(t=\sqrt{2})| = \sqrt{1^2 + (2\sqrt{2})^2} = \sqrt{1 + 8} = \sqrt{9} = 3 \neq 5 $$ A contradiction of sorts.
Point ($3$) above would lead one to naively assume that acceleration is $2$ units/sec$^2$ upwards for all times $t$. From there, one could conclude $\vec{F} = m \langle 0, 2 \rangle = \langle 0, 2m \rangle$. Alas the automated grader reports the correct answer is:
$$\displaystyle \vec{F} = \langle-\frac{-100\sqrt{2}}{81}m, \frac{50}{81}m\rangle$$
The connection between 3 (or 9) and 81 is not lost on me ($3^4 = 9^2 = 81$), but I don't see how to bridge the gap. Any help would be greatly appreciated. I've already got my grade and the "answer". What I seek now is understanding.
Many thanks in advance.