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Calculus/Particle Motion

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Question
A particle moves along the x-axis so that its velocity at any time t is given by v(t)= sin2t. At time t=0,the particle is at the origin.

a) For 0<or equal to t< or equal to pi, find the values of t for which that particle is moving to the left.

b) Write an expression for the position of the particle at any time t.

c) For 0<or equal to t< or equal to pi/2, find the average value of the position function determined in b.

Answer
Hi Reid,
So, we know that v = sin2t and x = 0 when t = 0.
a) If we use the convention where movement to the right indicates positive velocity and vice-versa, then the particle is moving to the left when v is negative.
sin2t < 0 when π ≤ 2t ≤ 2π
i.e π/2 ≤ t ≤ π

b) From v = dx/dt
dx = vdt
x = ∫v dt
= ∫sin2t dt
= -(1/2)cos2t + C
but x = 0 at t = 0
0 = -(1/2)cos(0) + C
0 = -1/2 + C
C = 1/2
So,
x = (1/2)cos2t + 1/2

c) For 0 ≤ t ≤ π/2, the average value of x is given by
1/(π/2 - 0) . ∫(1/2)cos2t + 1/2 dt (from 0 to π/2)
= (2/π) . [(1/4)sin2t + t/2] (from 0 to π/2)
= (2/π) . [((1/4)sinπ + π/4) - ((1/4)sin0 + 0/2)]
= (2/π) . [π/4 - 0]
= (2/π)(π/4)
= 1/2

Regards

Calculus

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