Dynamic question, please help?

Over a test, the acceleration of the down arrow linearly with distance s from its initial value of 16,000 m / s ^ 2 A to B, leaving zero after a trip of 24 "v is the speed of the arrow. Thanks,

This is not as easy as well I hope it will at least give a better response to any of your answers. acceleration = A = 16000 – 8000X where x (in feet) is in the range of 0 to 2 V = at = (16000 – 8000X) t (I assume that the initial rate of release is zero) acceleration in all distance dx = (16000 – 8000X) dx is the velocity at any point x is V (x) therefore the time required to travel on the arrow to any distance dx is dx / v (x), then the speed change distance dx = dv = (16000 – 8000X) dx / v So v.dv = (16000 – 8000X) dx integrating both sides v ^ 2 = 32000x – 8000X ^ 2 + 2K, where K is the constant of integration, in this case because v = 0 = 0 for x = 0 x = 2, v ^ 2 = 64,000 to 32,000 = 32000 therefore v = 179 fps CHECK: In fact, since the acceleration is linear in the distance you can use the average value of 8000 in the distance acceleration. This obviously impart the same energy to the arrow. S = 1 / 2. For t ^ 2 = sqrt (4 / 8000) = 22.36ms v = at = 8000 x 0.02236 = 179 fps.

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