*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.*

2 2 2 2

d(t) = ∫v(t)dt = 12∫t^{2} dt - 36∫t dt + 15∫dt

0 0 0 0

d(t) = (4t^{3} - 18t^{2} + 15t) evaluate at d(t=2) - d(t=0)

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