Kinematics, 4 questions An angler sits (stationary) on shore watching a river fl
ID: 1328821 • Letter: K
Question
Kinematics, 4 questions
An angler sits (stationary) on shore watching a river flow by, and notices a fish accelerating upstream at constant acceleration a. The fish's water speed (speed according to the fish not the angler) at time to = 0 is vo = 0. After accelerating for t1 = 7 s the fish has a water speed of v1 = 5 ft/s. At time t1 the fish swims past the angler's location. At this moment, according to the stationary angler, the fish is moving only v1 = 2.5 ft/s. Refer to Figure 1 and report the following: The speed of the river's current. The acceleration of the fish in the reference frame of the fish. The acceleration of the fish in the reference frame of the angler. Compare this to the answer from part (c). How far downstream the fish is from the angler's position on the bank at time to. The distance the angler would need to throw a rock in order to hit the fish at time tf = 10 s. (Assume the fish continues accelerating until struck by the rock.) Hint: This is now a 2-dimensional problem, refer to Fig. 1. Start by finding how far past the angler the fish has swam at timel tf. The velocity of the thrown rock as it leaves the angler's hand. The rock's trajectory is initially horizontal, and the rock leaves the angler's hand at waist height (ho = 3 ft above the level of the fish swimming at the surface of the river).Explanation / Answer
a.use relative speed in Opposite direction = diff of speerds
so speed of river = V1-V2 = 7-5 = 2 ft/s
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b. use the formula for accleration a = V/t
acleration of the fish = v/t = 2.5/7 = 0.357 ft/s^2
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c. when an angle is enclosed ,
Accleration (angle) = 2/7 = 0.285 ft/s^2
ratio af/af' = 0.357/0.285 = 1.25
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D.
we have to use the kinematic equation X = vot + 0.5 at^2
X = ( 5* 10) + ( 0.5* 0.357 * 10*10)
X = 67.85 ft
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e. tan theta = ay/ax
tan theta = 0.285/0.357
theta = 38.6 deg
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f. use V = sqrt(2gh)
V = sqrt(2* 32* 3)
V = 13.85 ft/s
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