Question

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 of6.00 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?

Short answer

a) 51.42 m.

b) 17.14 s.

Step-by-step solution

Given data

• Initial velocity U = 0.
• Final velocity V = 6.00 m/s .
• Acceleration of the swan a = 0.350 m/s².

Distance traveled by the swan

a) The distance traveled by the swan can be calculated using the equation as:

$V^2-U^2=2ad$

Here V is the final velocity, U is the initial velocity, a is the acceleration, and d is the distance traveled.

Substituting values in the above expression, we get,

$$\begin{gathered} (6{)}^2-(0{)}^2=2\times (0.35)\times d \\ 36=2\times (0.35)\times d \\ d=\frac{36}{2\times (0.35)} \\ d=51.42\text{m} \end{gathered}$$

The distance traveled by the bird is 51.42 m.

The time is taken by the swan

b) The time it takes for the swan to take-off can be calculated as:

V= U + at

Here V is the final velocity, U is the initial velocity, a is the acceleration, and t is the time.

$$\begin{gathered} 6=0+(0.35)\times t \\ t=\frac{6}{0.35} \\ t=17.14\text{s} \end{gathered}$$

Thus, it takes 17.14 s to take the flight.