College Physics 2.31 – Acceleration of a swan getting airborne


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

\text{v}_0=0\:\text{m/s}

\text{v}=6.00\:\text{m/s}

\text{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 \:\text{v}^2=\text{v}_0^2+2\text{a}\Delta \text{x}. Substituting the given values, we have

\left(6.00\:\text{m/s}\right)^2=\left(0\:\text{m/s}\right)^2+2\left(0.350\:\text{m/s}^2\right)\Delta x

\displaystyle \Delta x=\frac{\left(6.00\:\text{m/s}\right)^2}{2\left(0.350\:\text{m/s}^2\right)}

\Delta x=51.4286\:\text{m}

Part B

From the basic formula for acceleration, \text{a}=\frac{\text{v}-\text{v}_0}{\text{t}}, we have the formula for time:

\displaystyle \text{t}=\frac{\text{v}-\text{v}_0}{\text{a}}

\displaystyle \text{t}=\frac{6.00\:\text{m/s}-0\:\text{m/s}}{0.350\:\text{m/s}^2}

\text{t}=17.1429\:\sec