1. **Problem Statement:** A race car must achieve an average speed of 250 km/h over a 1600 m track. It covers the first half (800 m) at 230 km/h. Find the minimum average speed for the second half to qualify. 2. **Formula and Concepts:** Average speed $v_{avg}$ over total distance $d$ is total distance divided by total time: $$v_{avg} = \frac{d}{t_{total}}$$ Time for each half is distance divided by speed: $$t_1 = \frac{d_1}{v_1}, \quad t_2 = \frac{d_2}{v_2}$$ Total time: $$t_{total} = t_1 + t_2$$ 3. **Given:** - Total distance $d = 1600$ m - First half $d_1 = 800$ m, speed $v_1 = 230$ km/h - Second half $d_2 = 800$ m, speed $v_2 = ?$ - Required average speed $v_{avg} = 250$ km/h 4. **Convert speeds to m/s:** $$v_1 = 230 \times \frac{1000}{3600} = \frac{230000}{3600} \approx 63.89\ \mathrm{m/s}$$ $$v_{avg} = 250 \times \frac{1000}{3600} = \frac{250000}{3600} \approx 69.44\ \mathrm{m/s}$$ 5. **Calculate time for first half:** $$t_1 = \frac{800}{63.89} \approx 12.52\ \mathrm{s}$$ 6. **Total time allowed for whole track:** $$t_{total} = \frac{1600}{69.44} \approx 23.04\ \mathrm{s}$$ 7. **Time left for second half:** $$t_2 = t_{total} - t_1 = 23.04 - 12.52 = 10.52\ \mathrm{s}$$ 8. **Minimum speed for second half:** $$v_2 = \frac{d_2}{t_2} = \frac{800}{10.52} \approx 76.06\ \mathrm{m/s}$$ 9. **Convert back to km/h:** $$v_2 = 76.06 \times \frac{3600}{1000} = 273.8\ \mathrm{km/h}$$ **Final answer:** The car must average at least **273.8 km/h** in the second half to qualify. --- 1. **Problem Statement:** Two athletes run on a straight track. Athlete 1 runs at speed $v_1$ and is distance $d$ behind athlete 2 running at speed $v_2$. Find conditions and formulas for overtaking. 2. **(a) Condition to overtake:** Athlete 1 can overtake athlete 2 only if $v_1 > v_2$. 3. **(b) Time to overtake:** Let $t$ be time after start when athlete 1 catches athlete 2. Positions: $$x_1 = v_1 t$$ $$x_2 = d + v_2 t$$ Set $x_1 = x_2$: $$v_1 t = d + v_2 t \implies t = \frac{d}{v_1 - v_2}$$ 4. **(c) Minimum finish line distance $d_c$ for athlete 1 to tie:** Athlete 1 covers distance: $$x_1 = v_1 t = v_1 \frac{d}{v_1 - v_2}$$ Athlete 2 covers distance: $$x_2 = d + v_2 t = d + v_2 \frac{d}{v_1 - v_2} = d \frac{v_1}{v_1 - v_2}$$ To tie, finish line distance $d_c$ must be at least: $$d_c = x_2 = d \frac{v_1}{v_1 - v_2}$$ --- 1. **Problem Statement:** Find instantaneous velocity of a particle at times $t=1.00, 3.00, 4.50, 7.50$ s from a position vs. time graph. 2. **Method:** Instantaneous velocity is slope of tangent to position-time graph at given time. 3. **Approximate slopes from graph segments:** - At $t=1.00$ s, position near 10 m, slope positive and steep. - At $t=3.00$ s, position decreasing, slope negative. - At $t=4.50$ s, slope less steep negative. - At $t=7.50$ s, slope near zero or slightly negative. 4. **Exact values require graph data; conceptually:** Instantaneous velocity $v = \frac{\Delta x}{\Delta t}$ over small intervals around each time. --- 1. **Problem Statement:** Position of race car given by $$x = (5.0\ \mathrm{m/s}) t + (0.75\ \mathrm{m/s}^3) t^3$$ Find instantaneous velocity at $t=4.0$ s using time intervals. 2. **Velocity formula:** Instantaneous velocity is derivative: $$v = \frac{dx}{dt} = 5.0 + 3 \times 0.75 t^2 = 5.0 + 2.25 t^2$$ 3. **Calculate at $t=4.0$ s:** $$v = 5.0 + 2.25 \times 16 = 5.0 + 36 = 41.0\ \mathrm{m/s}$$ 4. **Using time intervals:** Calculate average velocity over intervals near 4.0 s, e.g., from 3.5 to 4.5 s: $$x(3.5) = 5.0 \times 3.5 + 0.75 \times 3.5^3 = 17.5 + 0.75 \times 42.875 = 17.5 + 32.156 = 49.656\ \mathrm{m}$$ $$x(4.5) = 5.0 \times 4.5 + 0.75 \times 4.5^3 = 22.5 + 0.75 \times 91.125 = 22.5 + 68.344 = 90.844\ \mathrm{m}$$ Average velocity: $$v_{avg} = \frac{90.844 - 49.656}{4.5 - 3.5} = \frac{41.188}{1.0} = 41.188\ \mathrm{m/s}$$ Close to derivative value 41.0 m/s. --- 1. **Problem Statement:** Two runners start at different positions and speeds towards a flagpole. Find distance from flagpole when they meet. 2. **Positions:** Runner A starts 4.0 mi west, velocity $+6.0$ mi/h east. Runner B starts 3.0 mi east, velocity $-5.0$ mi/h west. 3. **Set meeting point at distance $x$ east of flagpole:** Runner A position: $$x_A = -4.0 + 6.0 t$$ Runner B position: $$x_B = 3.0 - 5.0 t$$ Set $x_A = x_B$: $$-4.0 + 6.0 t = 3.0 - 5.0 t \implies 11 t = 7 \implies t = \frac{7}{11} \approx 0.636\ \mathrm{h}$$ 4. **Distance from flagpole:** $$x = x_A = -4.0 + 6.0 \times 0.636 = -4.0 + 3.82 = -0.18\ \mathrm{mi}$$ Negative means 0.18 mi west of flagpole. --- 1. **Problem Statement:** A particle accelerates as per Figure P2.20 with acceleration steps: 0 m/s² (0-5 s), 2 m/s² (5-15 s), -2 m/s² (15-20 s). Find speed at 10 s and 20 s, and distance traveled in 20 s. 2. **Calculate velocity:** - From 0 to 5 s, $a=0$, velocity constant at 0. - From 5 to 10 s (5 s interval), $a=2$, velocity change: $$\Delta v = a \Delta t = 2 \times 5 = 10\ \mathrm{m/s}$$ Velocity at 10 s: $$v = 0 + 10 = 10\ \mathrm{m/s}$$ - From 10 to 15 s (5 s), $a=2$, velocity change: $$\Delta v = 2 \times 5 = 10$$ Velocity at 15 s: $$v = 10 + 10 = 20\ \mathrm{m/s}$$ - From 15 to 20 s (5 s), $a=-2$, velocity change: $$\Delta v = -2 \times 5 = -10$$ Velocity at 20 s: $$v = 20 - 10 = 10\ \mathrm{m/s}$$ 3. **Distance traveled:** Calculate displacement in each interval using average velocity: - 0-5 s: velocity 0, displacement 0. - 5-15 s: velocity changes from 0 to 20 m/s, average 10 m/s, time 10 s: $$d = 10 \times 10 = 100\ \mathrm{m}$$ - 15-20 s: velocity changes from 20 to 10 m/s, average 15 m/s, time 5 s: $$d = 15 \times 5 = 75\ \mathrm{m}$$ Total distance: $$0 + 100 + 75 = 175\ \mathrm{m}$$ --- 1. **Problem Statement:** A train moving at 20 m/s brakes with acceleration -1.0 m/s² for 40 s. Find distance traveled during braking. 2. **Use kinematic equation:** $$d = v_0 t + \frac{1}{2} a t^2$$ $$d = 20 \times 40 + \frac{1}{2} (-1.0) \times 40^2 = 800 - 800 = 0\ \mathrm{m}$$ Velocity after 40 s: $$v = v_0 + a t = 20 - 40 = -20\ \mathrm{m/s}$$ Negative velocity means train reversed direction, so it stopped before 40 s. 3. **Find stopping time:** $$0 = 20 - 1.0 t \implies t = 20\ \mathrm{s}$$ Distance to stop: $$d = 20 \times 20 + \frac{1}{2} (-1.0) \times 20^2 = 400 - 200 = 200\ \mathrm{m}$$ 4. **Distance traveled in 40 s:** Train stops at 20 s, then moves backward 20 s at increasing speed. Total displacement after 40 s is 0 m. --- 1. **Problem Statement:** A ball is thrown vertically upward at 25.0 m/s. Find (a) max height, (b) time to max height, (c) time to hit ground after max height, (d) velocity on return. 2. **Use kinematic equations:** Acceleration $a = -9.8$ m/s². 3. **(a) Max height:** $$v = v_0 + a t = 0 \implies t = \frac{-v_0}{a} = \frac{-25.0}{-9.8} = 2.55\ \mathrm{s}$$ Height: $$h = v_0 t + \frac{1}{2} a t^2 = 25.0 \times 2.55 - 4.9 \times (2.55)^2 = 31.9\ \mathrm{m}$$ 4. **(b) Time to max height:** $2.55$ s 5. **(c) Time to hit ground after max height:** Same as time up, $2.55$ s 6. **(d) Velocity on return:** Equal magnitude, opposite direction: $$v = -25.0\ \mathrm{m/s}$$ --- 1. **Problem Statement:** A 50.0-g ball hits a wall at 25.0 m/s and rebounds at 22.0 m/s. Contact time 3.50 ms. Find average acceleration magnitude. 2. **Change in velocity:** $$\Delta v = v_{final} - v_{initial} = (-22.0) - 25.0 = -47.0\ \mathrm{m/s}$$ 3. **Average acceleration:** $$a = \frac{\Delta v}{\Delta t} = \frac{-47.0}{3.50 \times 10^{-3}} = -13429\ \mathrm{m/s}^2$$ Magnitude: $$|a| = 1.34 \times 10^4\ \mathrm{m/s}^2$$ --- 1. **Problem Statement:** A car accelerates uniformly from 55 mi/h to 60 mi/h at 0.60 m/s². Find time taken. 2. **Convert speeds to m/s:** $$55 \times \frac{1609}{3600} = 24.59\ \mathrm{m/s}$$ $$60 \times \frac{1609}{3600} = 26.82\ \mathrm{m/s}$$ 3. **Time:** $$t = \frac{v_f - v_i}{a} = \frac{26.82 - 24.59}{0.60} = 3.72\ \mathrm{s}$$