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

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