# College Physics 2.31 – Acceleration of a swan getting airborne

## 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:

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

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

$\text{t}=17.1429\:\sec$