College Physics by Openstax Chapter 2 Problem 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/s2, how far will it travel before becoming airborne?

(b) How long does this take?

Solution:

Part A

We are given the following values: vf=6.00 m/sv_f=6.00\:\text{m/s}; v0=0 m/sv_0=0\:\text{m/s}; and a=0.350 m/s2a=0.350\:\text{m/s}^2.

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

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

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

Δx=(vf)2−(v0)22a\Delta x=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2a}

Substituting the given values,

Δx=(vf)2−(v0)22aΔx=(6.00 m/s)2−(0.00 m/s)22(0.350 m/s2)Δx=51.4286 m  (Answer)\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 vf=v0+atv_f=v_0+at, solve for time, tt in terms of the other variables.

t=vf−v0at=\frac{v_f-v_0}{a}

Substitute the given values

t=vf−v0at=6.00 m/s−0.00 m/s0.350 m/s2t=17.1429 s  (Answer)\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*}
Advertisements
Advertisements