A force with magnitude N acts directly upward from the xy-plane on an object with mass kg. The object starts at the origin with initial velocity . Find its position function.

Question

A force with magnitude  N acts directly upward from the xy-plane on an object with mass  kg. The object starts at the origin with initial velocity  . Find its position function.

Quizplus · Accepted Answer

Using Newton's second law, we have
\begin{align*}
F &= ma \
11eaa3e5\_54ef\_7a02\_9f8f\_dbfd099b849a\_TB5971\_11 &= 11eaa3e5\_54ef\_7a03\_9f8f\_e140cff88af2\_TB5971\_11 \frac{d^2\vec{r}}{dt^2} \
\end{align*}
Since the force is acting directly upward, we can write $\vec{F}=(0, 0, 11eaa3e5\_54ef\_7a02\_9f8f\_dbfd099b849a\_TB5971\_11)$ in Cartesian coordinates. Since the problem is symmetrical with respect to the $z$-axis, we can assume without loss of generality that the motion lies in the $xz$-plane, so that $\vec{r}=(x,0,z)$.
Using Cartesian coordinates for the position vector, we have
\begin{align*}
\frac{d^2\vec{r}}{dt^2} &= \frac{d^2}{dt^2}(x,0,z) \
&= \left(\frac{d^2x}{dt^2},0,\frac{d^2z}{dt^2}ight)
\end{align*}
Therefore, we have
\begin{align*}
\left(0,0,11eaa3e5\_54ef\_7a02\_9f8f\_dbfd099b849a\_TB5971\_11ight) &= 11eaa3e5\_54ef\_7a03\_9f8f\_e140cff88af2\_TB5971\_11 \left(\frac{d^2x}{dt^2},0,\frac{d^2z}{dt^2}ight) \
&= 11eaa3e5\_54ef\_7a03\_9f8f\_e140cff88af2\_TB5971\_11 \frac{d^2}{dt^2}(x,0,z) \
&= 11eaa3e5\_54ef\_7a03\_9f8f\_e140cff88af2\_TB5971\_11 \frac{d^2\vec{r}}{dt^2}
\end{align*}
which implies
\begin{align*}
\frac{d^2x}{dt^2} &= 0 \
\frac{d^2z}{dt^2} &= 11eaa3e5\_54ef\_7a02\_9f8f\_dbfd099b849a\_TB5971\_11 / 11eaa3e5\_54ef\_7a03\_9f8f\_e140cff88af2\_TB5971\_11 \
&= 11eaa3e5\_54ef\_7a00\_9f8f\_fa4df73f9974\_TB5971\_11 \
&= 1.1 \ 	ext{m}/	ext{s}^2
\end{align*}
Integrating once with respect to time, we have
\begin{align*}
\frac{dx}{dt} &= c_1 \
\frac{dz}{dt} &= 1.1t + c_2
\end{align*}
where $c_1$ and $c_2$ are constants of integration.
Using the initial condition $\vec{r}(0)=(0,0,0)$, we have $\frac{dx}{dt}(0)=0$ and $\frac{dz}{dt}(0)=0$. Therefore, $c_1=0$ and $c_2=0$, and we get
\begin{align*}
x(t) &= 0 \
z(t) &= \frac{1}{2}(1.1t)^2 \
&= 0.605t^2
\end{align*}
Thus, the position function is $\vec{r}=\left(0,0,\frac{1}{2}(1.1t)^2ight)$, which matches choice (B).

A Force with Magnitude N Acts Directly Upward from the Xy-Plane