An object moves along a straight line with acceleration a(t)=3/ sqrt (t) for t >= 1.
If it's velocity at t=4 is10, find its location at t=16.
acceleration, velocity.. 15pts
2009-11-06 7:24 am
回答 (1)
2009-11-06 7:45 am
✔ 最佳答案
a = 3t^(-1/2)v = ∫adt = 6√t + C where C is a constant
when t = 4, v = 10
10 = 12 + C => C = -2
v = 6√t - 2
s = ∫vdt = 4t√t - 2t + K where K is a constant
when t = 1, s = 0 (consider the position relative to t = 1)
so 4 - 2 + K = 0 => k = -2
s = 4t√t - 2t - 2
At t = 16, s= 4*16*4 - 2*4 - 2 = 246
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