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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.

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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