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Expert-verified Found in: Page 35 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # 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?

a) 51.42 m.

b) 17.14 s.

See the 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,

$\left(6{\right)}^{2}-\left(0{\right)}^{2}=2×\left(0.35\right)×d\phantom{\rule{0ex}{0ex}}36=2×\left(0.35\right)×d\phantom{\rule{0ex}{0ex}}d=\frac{36}{2×\left(0.35\right)}\phantom{\rule{0ex}{0ex}}d=51.42\text{m}$

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.

$6=0+\left(0.35\right)×t\phantom{\rule{0ex}{0ex}}t=\frac{6}{0.35}\phantom{\rule{0ex}{0ex}}t=17.14\text{s}$

Thus, it takes 17.14 s to take the flight. ### Want to see more solutions like these? 