The acceleration of an object moving along the x-axis is given by a(t)=18t-2, where the velocity is 12 when t=2 and the position is 2 when t=1. The position x(t) =
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Okay, if the position is defined to be x(t), then the velocity v(t) is dx/dt and the acceleration a(t) (conveniently provided) is d^2x/dt^2 (i.e., x''(t), the second derivative). Integrating a(t) with respect to t, we get
v(t) = 9t^2 -2t + c (note that c is the starting velocity, v(0))
Luckily, we are given the velocity at a specified point:
v(2) = 12 = (9 * 4) - 4 + c, so
12 - 32 = -20 = c.
Thus, v(t) = 9t^2 -2t -20
(I'm using the star (*) operator for multiplication BTW.)
One more antiderivative:
v(t) = x'(t), so x(t) = (1/3)(9t^3) -t^2 - 20t + x(0)
= 3t^3 - t^2 - 20t + x(0)
Given that x(1) = 2, this means that 3 - 1 - 20 + x(0) = 2; so x(0) = 18. Thus,
x(t) = 3t^3 - t^2 - 20t + 18
Aren't parametric equations wonderful?
v(t) = 9t^2 -2t + c (note that c is the starting velocity, v(0))
Luckily, we are given the velocity at a specified point:
v(2) = 12 = (9 * 4) - 4 + c, so
12 - 32 = -20 = c.
Thus, v(t) = 9t^2 -2t -20
(I'm using the star (*) operator for multiplication BTW.)
One more antiderivative:
v(t) = x'(t), so x(t) = (1/3)(9t^3) -t^2 - 20t + x(0)
= 3t^3 - t^2 - 20t + x(0)
Given that x(1) = 2, this means that 3 - 1 - 20 + x(0) = 2; so x(0) = 18. Thus,
x(t) = 3t^3 - t^2 - 20t + 18
Aren't parametric equations wonderful?