Question: A particle moves according to a law of motion s = f(t), t \geq 0, where t is measured in seconds and s in feet. f(t) = t^3 - 15t^2 + 72t (a) Find the velocity at time t (in ft/s). v(t) = (b) What is the velocity (in ft/s) after 5 s? v(5) = ft/s (c) When (in seconds) is the particle at rest? (smaller value) t = s (larger value) t = s (d) When (in seconds) is the particle moving in the positive direction? (Enter your answer using interval notation.) t is in (e) Find the total distance (in feet) traveled during the first 8 s. ft (f) Find the acceleration at time t (in ft/s^2). a(t) = Find the acceleration (in ft/s^2) after 5 s. a(5) = ft/s^2 (g) Graph the position, velocity, and acceleration functions for the first 8 s. (h) When, for 0 \leq t < \infty, is the particle speeding up? (Enter your answer using interval notation.) When, for 0 \leq t < \infty, is it slowing down? (Enter your answer using interval notation.)

1. **State the problem:** We have a particle moving with position function $$f(t) = t^3 - 15t^2 + 72t$$ for $$t \geq 0$$, where $$t$$ is time in seconds and $$s$$ is position in feet. 2. **Find velocity function $$v(t)$$:** Velocity is the derivative of position with respect to time: $$v(t) = f'(t) = \frac{d}{dt}(t^3 - 15t^2 + 72t) = 3t^2 - 30t + 72$$ 3. **Find velocity at $$t=5$$:** $$v(5) = 3(5)^2 - 30(5) + 72 = 3(25) - 150 + 72 = 75 - 150 + 72 = -3$$ ft/s 4. **Find when particle is at rest:** Particle is at rest when velocity is zero: $$3t^2 - 30t + 72 = 0$$ Divide entire equation by 3: $$\cancel{3}t^2 - \cancel{30}t + \cancel{72} = 0 \Rightarrow t^2 - 10t + 24 = 0$$ Factor quadratic: $$(t - 6)(t - 4) = 0$$ So, $$t = 4, 6$$ seconds 5. **Find when particle moves in positive direction:** Velocity $$v(t) = 3t^2 - 30t + 72$$ is positive. From roots $$t=4$$ and $$t=6$$, test intervals: - For $$t < 4$$, pick $$t=0$$: $$v(0) = 72 > 0$$ positive - For $$4 < t < 6$$, pick $$t=5$$: $$v(5) = -3 < 0$$ negative - For $$t > 6$$, pick $$t=7$$: $$v(7) = 3(49) - 210 + 72 = 147 - 210 + 72 = 9 > 0$$ positive Thus, particle moves positively on intervals $$[0,4) \cup (6, \infty)$$ 6. **Find total distance traveled in first 8 seconds:** Calculate position at $$t=0,4,6,8$$: $$f(0) = 0$$ $$f(4) = 64 - 240 + 288 = 112$$ $$f(6) = 216 - 540 + 432 = 108$$ $$f(8) = 512 - 960 + 576 = 128$$ Distance traveled: From 0 to 4: $$|112 - 0| = 112$$ From 4 to 6: $$|108 - 112| = 4$$ From 6 to 8: $$|128 - 108| = 20$$ Total distance = $$112 + 4 + 20 = 136$$ feet 7. **Find acceleration function $$a(t)$$:** Acceleration is derivative of velocity: $$a(t) = v'(t) = \frac{d}{dt}(3t^2 - 30t + 72) = 6t - 30$$ 8. **Find acceleration at $$t=5$$:** $$a(5) = 6(5) - 30 = 30 - 30 = 0$$ ft/s² 9. **Speeding up and slowing down:** Speeding up when velocity and acceleration have the same sign. Slowing down when velocity and acceleration have opposite signs. - Velocity roots: $$t=4,6$$ - Acceleration root: $$a(t) = 0 \Rightarrow 6t - 30 = 0 \Rightarrow t=5$$ Test intervals: - $$0 \leq t < 4$$: $$v>0$$, $$a=6t-30<0$$ (since $$6(0)-30=-30<0$$), opposite signs → slowing down - $$4 < t < 5$$: $$v<0$$, $$a<0$$ same signs → speeding up - $$5 < t < 6$$: $$v<0$$, $$a>0$$ opposite signs → slowing down - $$t > 6$$: $$v>0$$, $$a>0$$ same signs → speeding up **Intervals:** - Speeding up: $$(4,5) \cup (6, \infty)$$ - Slowing down: $$[0,4) \cup (5,6)$$