# College Physics Problem 2.31

#### 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  $0.350\:m/s^2$, how far will it travel before becoming airborne? (b) How long does this take?

SOLUTION:

Part a.
We are given

$v_0=0$
$v=6.00\:m/s$
$a=0.350\:m/s^2$

From the kinematical equations, the most applicable formula to solve for the change in distance, $\Delta x$, is $v^2=v_0^2+2a\Delta x$. Substituting the given values, we have

$\left(6.00\:m/s\right)^2=\left(0\right)^2+2\left(0.350\:m/s^2\right)\Delta x$
$\Delta \:x=\frac{\left(6.00\:m/s\right)^2}{2\left(0.350\:m/s^2\right)}$
$\Delta \:x=51.4286\:m/s^2$

Part b
From the basic formula for acceleration, $a=\frac{v-v_0}{t}$, we have the formula for time:

$t=\frac{v-v_0}{a}$
$t=\frac{6.00\:m/s}{0.350\:m/s^2}$
$t=17.1429\:sec$

You can now buy the complete solution manual for College Physics. Just complete the google form below.