Findinh total distance via Velocity and Determining when a particle speeds up and slows down.

Question

a particle moves a long a straight line and its position at time t is given by s(t)= t³-10.5²+30t where s is measured in feet and t in seconds. Consider t in the interval [0,infinity).

1. what is the total distance the particle travels between 0 and time 7?
2. when is the particle speeding up? (enter as an interval)
3. when is the particle slowing down?

Answer 1

s(t) = t³ - 10.5t² + 30t

s'(t) = 3t² - 21t + 30 = 3(t² - 7t + 10) = 3(t - 5)(t - 2)

s"(t) = 6t - 21

s'(t) = 0 when t = 2 or t = 5

s"(t) = 0 when t = 3.5

When 0 < t < 2, s'(t) > 0, So, the particle is moving to the right.

When 2 < t < 5, s'(t) < 0. So, the object is moving to the left.

When 5 < t < ∞, s'(t) > 0. So, the object is moving to the right.

Total distance traveled from t = 0 to t = 7 is: [s(2) - s(0)] + [s(2) - s(5)] + [s(7) - s(5)]

= [86 - 0] + [86 - 12.5] + [38.5 - 12.5] = 185.5 ft

When 0 < t < 3.5, s"(t) < 0 and when t > 3.5, s"(t) > 0.

The particle speeds up when s'(t) and s"(t) have the same sign . This occurs when 2 < t < 3.5 and when

t > 5. The particle slows down when s'(t) and s"(t) have opposite signs. This happens when 0 < t < 2 and when 3.5 < t < 5.

Answer 2

s(t)=t^3-10.5^2+30t

d=343-110.25+210=442.75