A swan on a lake gets airborne by its wing and running on top of the water. (a)

Question

A swan on a lake gets airborne by its wing and running on top of the water. (a) If the swan must reach a velocity of 5.55 m/s to take off and it accelerates from rest at an average rate of 0.300 m/s^2, how far (in m) will it travel before m (b) How long (in s) does this take? s In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no to escape certain death. Some about 20, 177 feet (6150 m), and some of them survived, with few life threatening injuries. For these lucky the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot's speed upon impact was 114 mph (51 m/s), then what was his deceleration (in m/s^2)? Assume that the trees and snow stopped him over a distance of 3.9 magnitude.) m/s^2

Explanation / Answer

a)

v = u + at

here , v = 5.55 m/s

a = 0.3 m/s2

u = 0 m/s

So, 5.55 = 0 + 0.3*t

So, t = 18.5 s

So, distance traveled = s = ut + 0.5*at^2

= 0 + 0.5*5.55*18.5^2 = 949.7 m <------- answer

b)

time taken , t = 18.5 s <----- answer

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