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Chapter 2: Q2.5-31PE (page 35)

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

Expert verified

a) 51.42 m.

b) 17.14 s.

Step by step solution

01

Given data

  • Initial velocity U = 0.
  • Final velocity V = 6.00 m/s .
  • Acceleration of the swan a = 0.350 m/s2.
02

Distance traveled by the swan

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

V2-U2=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×(0.35)×d36=2×(0.35)×dd=362×(0.35)d=51.42m

The distance traveled by the bird is 51.42 m.

03

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+(0.35)×tt=60.35t=17.14s

Thus, it takes 17.14 s to take the flight.

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