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 t2 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∫t2 dt - 36∫t dt + 15∫dt
0 0 0 0
d(t) = (4t3 - 18t2 + 15t) evaluate at d(t=2) - d(t=0)