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Calc. 3 problem!!!?/?
Suppose that the acceleration is given by a(t) = <cos t; 2 sin t>, the
initial position is given by <1, 0> and the initial velocity by <0, 1>. Then
fi nd the vector valued functions that give the position and the velocity.
please give details
1 Answer
- Anonymous9 years agoFavorite Answer
v(t) = ∫ a(t) dt
v(t) = ∫ <cos(t), 2sin(t)> dt
v(t) = <sin(t) + C1, -2cos(t) + C2>
Since the initial velocity (which means v(t) at t = 0) is <0, 1>:
sin(0) + C1 = 0
0 + C1 = 0
C1 = 0, and
-2cos(0) + C2 = 1
-2(1) + C2 = 1
C2 = 3
So the velocity is
v(t) = <sin(t), -2cos(t) + 3>
s(t) = ∫ v(t) dt
s(t) = ∫ <sin(t), -2cos(t) + 3> dt
s(t) = <-cos(t) + C3, -2sin(t) + 3t + C4>
Again, the initial position, x(0), is given as <1, 0>
-cos(0) + C3 = 1
-1 + C3 = 1
C3 = 2
-2sin(0) + 3(0) + C4 = 0
-2(0) + 0 + C4 = 0
C4 = 0
Thus:
s(t) = <-cos(t) + 2, -2sin(t) + 3t>