Aswan 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?

The distance is  51.42 meters.

The time is  17.14 seconds approximately.

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

The time is  17.14 seconds approximately.