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 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?

Answer 1

The swan will travel approximately 102.86 meters before becoming airborne and it will take approximately 17.14 seconds for the swan to become airborne.

Step-by-step explanation:

The distance a swan will travel before becoming airborne can be calculated using the equation:

d = (v^2 - u^2) / (2a)

Where:

• d is the distance traveled
• v is the final velocity (6.00 m/s)
• u is the initial velocity (0 m/s)
• a is the acceleration (0.350 m/s²)

Plugging in the given values:

d = (6.00^2 - 0^2) / (2 * 0.350)

d = 72.00 / 0.700

d ≈ 102.86 m

Therefore, the swan will travel approximately 102.86 meters before becoming airborne.

To calculate the time it takes for the swan to become airborne, we can use the equation:

t = (v - u) / a

Where:

• t is the time
• v is the final velocity (6.00 m/s)
• u is the initial velocity (0 m/s)
• a is the acceleration (0.350 m/s²)

Plugging in the given values:

t = (6.00 - 0) / 0.350

t = 6.00 / 0.350

t ≈ 17.14 s

Therefore, it will take approximately 17.14 seconds for the swan to become airborne.

Answer 2

The distance is 51.42 meters.

The time is 17.14 seconds approximately.

Step-by-step explanation:

from the following relation.

vf^2 = vi^2 + 2 * a * d

Where vf is final velocity

a is acce;eration

d is distance to which swan traveled

vi = 0 { accelerate from rest}

6.00^2 = 2 * 0.350 * d

d =\frac{36}{.7}

The distance is 51.42 meters.

(b)

Use the following equation.

d = ½ * (vi + vf) * t

51.42= ½ * 6.0 * t
`t = (51.42*2)/(6)`

The time is 17.14 seconds approximately.