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Q2.5-31PE

Expert-verifiedFound in: Page 35

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/s ^{2}**

**(b) How long does this take?**

a) 51.42 m.

b) 17.14 s.

- Initial velocity U =
**0**. - Final velocity V =
**6.00 m/s .** - Acceleration of the swan a =
**0.350 m/s**^{2}.

**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,

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

The distance traveled by the bird is 51.42 m.

**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.**

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

Thus, it takes 17.14 s to take the flight**.**

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