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Question

Aparticle travels horizontally so that its position at any time t on the closed interval [pi/2,5pi/2] is given by x(t) = (3t/pi )+cos(t). Find the particles position when it's acceleration is -1

Answer 1

x(t) = (3t/pi )+cos(t) Find the particles position when the acceleration is -1

x'(t) = 3/pi - sin(t) Velocity is the derivative of the distance.

x"(t) = - cos(t) Acceleration is the 2nd derivative of the distance.

When the acceleration is -1, t=0

Then the distance x(0) = 0 + cos(0)

x(0) = 1

Answer 2

First you have to find the acceleration of the particle by deriving x(t) with respect to t, twice. Then, solve for t at a = -1. Finally, plug the answer for t into the original equation of x.

1. v(t) = dx/dt = (3/π) - sin(t)
2. a(t) = dv/dt = -cos(t)
3. a(t) = 0 @ t = 2nπ s, were n is any integer.
4. x(t) @ t = nπ/2 = 1, 7, 13, 19, ...
5. Only x(t) = 7 falls withen [π/2, 5π/2], so the answer is x(t) = 7 when a = -1
6. Answer:
7. x(t) = 7 @ a(t) = -1