A particle P moves with constant acceleration (2i-3j) ms^-2. At time t seconds, its velocity is v ms^-1. When t=0, v=-2i+7j.
(a) Find the value of t when P is moving parallel to the vector i.
(b) Find the speed of P when t=3.
(c) Find the angle between the vector j and the direction of motion of P when t=3.
Question vector!~
2007-10-28 11:41 am
回答 (1)
2007-10-28 12:00 pm
✔ 最佳答案
Denote a = 2i-3ja.) v+at = Ai+0j
-2i+7j + 2t i - 3t j = Ai+0j
Comparing coefficient,
t=7/3 #
b.) v(3) = v+3a = 4i-2j |v| = sqrt(20)
c.) (4i-2j). j = |4i-2j| |j| cos Theta [dot product]
-2 = sqrt(20) cos Theta
and so on!
參考: me
收錄日期: 2021-04-23 20:07:18
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