1. Problem: A particle's position along a line is given by $s(t) = t^3 - 6t^2 - 15t + 7$ ft. Find the total distance traveled by $t=10$ s, the average velocity, average speed, and instantaneous velocity and acceleration at $t=10$ s. 2. Problem: Position of a particle is $s(t) = 1.5t^3 - 13.5t^2 + 22.5t$ ft. Find position at $t=6$ s and total distance traveled during this interval. 3. Problem: An airplane's velocity with drag parachute is $v(t) = \frac{80}{1+0.32t}$ m/s. Find acceleration at $t=3$ s. 4. Problem: A motorcycle starts from rest with velocity graph segments: $v=1.25t$ (0-4 s), $v=5$ (4-10 s), $v=-t+15$ (10-15 s). Find total distance until stopping at 15 s and plot acceleration graph. 5. Problem: A car traveling at 70 km/h accelerates at 6000 km/h$^2$. Find time to reach 120 km/h and distance traveled during acceleration. --- 1. Total distance requires checking for changes in direction by finding velocity $v(t) = s'(t) = 3t^2 - 12t - 15$. Solve $v(t)=0$: $$3t^2 -12t -15=0 \Rightarrow t^2 -4t -5=0 \Rightarrow t=5 \text{ or } t=-1$$ Ignore $t=-1$ since $t\ge0$. At $t=0$, $s=7$ ft; at $t=5$, $s(5)=125-150-75+7 = -93$ ft; at $t=10$, $s(10)=1000 -600 -150 +7=257$ ft. Distance segments: from 7 to -93 is $100$ ft; from -93 to 257 is $350$ ft. Total distance = $100 + 350 = 450$ ft. Average velocity: $$\text{avg velocity} = \frac{s(10)-s(0)}{10-0} = \frac{257-7}{10} = 25\text{ ft/s}.$$ Average speed is total distance over time: $$\text{avg speed} = \frac{450}{10} = 45\text{ ft/s}.$$ Instantaneous velocity and acceleration at $t=10$: $$v(10) = 3(10)^2 - 12(10) - 15 = 300 - 120 - 15 = 165\ \text{ft/s}.$$ $$a(t) = v'(t) = s''(t) = 6t -12.$$ $$a(10) = 6(10) - 12 = 60 - 12 = 48\ \text{ft/s}^2.$$ 2. Position at $t=6$ s: $$s(6) = 1.5(6)^3 -13.5(6)^2 + 22.5(6) = 1.5(216) - 13.5(36) + 135 = 324 - 486 + 135 = -27\ \text{ft}.$$ Velocity for distance : $$v(t) = s'(t) = 4.5t^2 - 27t + 22.5.$$ Solve $v(t) = 0$ on $[0,6]$: $$4.5t^2 - 27t + 22.5=0 \Rightarrow t^2 - 6t + 5=0 \Rightarrow t=1, 5.$$ Calculate position at these times: $$s(0) = 0,$$ $$s(1) = 1.5 -13.5 + 22.5 = 10.5,$$ $$s(5) = 1.5(125) - 13.5(25) + 22.5(5) = 187.5 - 337.5 + 112.5 = -37.5,$$ $$s(6) = -27.$$ Distance intervals: $0 \to 1:$ distance = $|10.5 - 0| = 10.5$ ft $1 \to 5:$ distance = $| -37.5 - 10.5| = 48$ ft $5 \to 6:$ distance = $| -27 - (-37.5)| = 10.5$ ft Total distance = $10.5 + 48 + 10.5 = 69$ ft. 3. Given velocity: $$v = \frac{80}{1 + 0.32t}.$$ Acceleration: $$a = \frac{dv}{dt} = -80 \times \frac{0.32}{(1+0.32t)^2} = -\frac{25.6}{(1 + 0.32t)^2}.$$ At $t = 3$: $$a(3) = -\frac{25.6}{(1 + 0.96)^2} = -\frac{25.6}{(1.96)^2} = -\frac{25.6}{3.8416} \approx -6.67\ \text{m/s}^2.$$ 4. Separate time intervals: - 0 to 4 s: $v = 1.25t$ - 4 to 10 s: $v = 5$ - 10 to 15 s: $v = -t + 15$ Distance traveled: $$\text{from } 0 \to 4: \int_0^4 1.25t dt = 1.25 \times \frac{4^2}{2} = 1.25 \times 8 = 10 \text{ m}.$$ $$\text{from } 4 \to 10: 5 \times (10-4) = 30 \text{ m}.$$ $$\text{from } 10 \to 15: \int_{10}^{15} (-t + 15) dt = \left[-\frac{t^2}{2} + 15t\right]_{10}^{15} =\left(-\frac{225}{2} + 225\right) - \left(-\frac{100}{2} + 150\right) = ( -112.5 + 225) - (-50 +150) = 112.5 - 100 = 12.5 \text{ m}.$$ Total distance: $$10 + 30 + 12.5 = 52.5 \text{ m}.$$ Acceleration $a = dv/dt$: - $0 \le t \le 4: a = 1.25$ m/s$^2$ - $4 < t < 10: a = 0$ - $10 \le t \le 15: a = -1$ m/s$^2$ 5. Initial velocity $v_0 = 70$ km/h, acceleration $a = 6000$ km/h$^2$, final velocity $v = 120$ km/h. Time to reach $120$: $$t = \frac{v - v_0}{a} = \frac{120 - 70}{6000} = \frac{50}{6000} = 0.008333... \text{ h} = 0.5 \text{ min} = 30 \text{ s}.$$ Distance traveled: $$s = v_0 t + \frac{1}{2} a t^2 = 70 \times 0.008333 + \frac{1}{2} \times 6000 \times (0.008333)^2 = 0.5833 + 0.2083 = 0.7916 \text{ km} = 791.6 \text{ m}.$$