A swan on a lake gets airborne by flapping its wings and running on top of the water.

(a) If the swan must reach a velocity of 6.00 m/s to take off and it accelerates  from rest at an average rate of  0.350 m/s², how far will it travel before becoming airborne?

(b) How long does this take?

Solution:

Part A

We are given the following values: v_f=6.00\:\text{m/s}; v_0=0\:\text{m/s}; and a=0.350\:\text{m/s}^2.

From the kinematic equations, the most applicable formula to solve for the change in distance, \Delta \text{x}, is

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

Solving for \Delta \text{x} in terms of the other variables, we have

\Delta x=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}

Substituting the given values,

\begin{align*}
\Delta x & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a} \\
\Delta x  & =\frac{\left(6.00\:\text{m/s}\right)^2-\left(0.00\:\text{m/s}\right)^2}{2\left(0.350\:\text{m/s}^2\right)} \\
\Delta x  & =51.4286\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

From the formula v_f=v_0+at, solve for time, t in terms of the other variables.

t=\frac{v_f-v_0}{a}

Substitute the given values

\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{6.00\:\text{m/s}-0.00\:\text{m/s}}{0.350\:\text{m/s}^2} \\
t & =17.1429\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}