A particle moves on the x-axis so that its velocity at any time t≥0 is given by v(t)=12t^2-36t+15. At t=1, the particle is at the origin.

a) Find the maximum velocity of the particle for 0≤t≤2.

a(t) = v'(t) = 24t-36 = 0 when t = 3/2
v(t=3/2) = -12

This is the minimum. Since coefficient of t^{2} term is positive (12), parabola opens upward with vertex at t = 3/2. Max value will be at t=0 or t = 2:

v(t=0) = 15 (max)

v(t=2) = -9

b) Find the total distance traveled by the particle from t=0 to t=2.

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A particle moves on the x-axis so that its velocity at any time t≥0 is given by v(t) = 12t^2 - 36t + 15. At t=1, the particle is at the origin.

a) Find the maximum velocity of the particle for 0≤t≤2.

v(t) is a parabola that opens up because the leading coefficient is positive. So the vertex is a minimum, not a maximum. Let’s find it, though:
h = 36/24 = 3/2, k = 15 - 12(3/2)^2 = 15-27 = -12
Vertex = (3/2,-12)

The maximum on the interval 0≤t≤2 must occur at one of the endpoints:

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