1. **State the problem:** Given the displacement function $s(t) = t^3 - 6t^2 - 36t$, find the maximum speed and maximum velocity in the first 7 seconds of motion. 2. **Recall formulas:** Velocity $v(t)$ is the first derivative of displacement: $$v(t) = \frac{ds}{dt}$$ Speed is the absolute value of velocity: $$\text{speed} = |v(t)|$$ 3. **Find velocity:** Differentiate $s(t)$: $$v(t) = \frac{d}{dt}(t^3 - 6t^2 - 36t) = 3t^2 - 12t - 36$$ 4. **Find critical points for velocity:** Set $v(t) = 0$ to find when velocity changes: $$3t^2 - 12t - 36 = 0$$ Divide both sides by 3: $$\cancel{3}t^2 - \cancel{12}t - \cancel{36} = 0 \Rightarrow t^2 - 4t - 12 = 0$$ 5. **Solve quadratic:** $$t = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times (-12)}}{2} = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2}$$ So, $$t_1 = \frac{4 + 8}{2} = 6, \quad t_2 = \frac{4 - 8}{2} = -2$$ Only $t=6$ is in the interval $[0,7]$. 6. **Evaluate velocity at critical points and endpoints:** $$v(0) = 3(0)^2 - 12(0) - 36 = -36$$ $$v(6) = 3(6)^2 - 12(6) - 36 = 3(36) - 72 - 36 = 108 - 108 = 0$$ $$v(7) = 3(7)^2 - 12(7) - 36 = 3(49) - 84 - 36 = 147 - 120 = 27$$ 7. **Find maximum velocity:** Among $v(0) = -36$, $v(6) = 0$, and $v(7) = 27$, the maximum velocity is $27$ at $t=7$ seconds. 8. **Find maximum speed:** Speed is $|v(t)|$. Check speed at $t=0,6,7$: $$|v(0)| = 36, \quad |v(6)| = 0, \quad |v(7)| = 27$$ Maximum speed is $36$ at $t=0$ seconds. **Final answers:** - Maximum velocity in first 7 seconds is $27$ at $t=7$. - Maximum speed in first 7 seconds is $36$ at $t=0$.