1. **Problem:** An automobile accelerates from rest at $1\ \text{m/s}^2$ for 30 s, continues at constant speed for 2 min, and then comes to a stop in 15 s. Find the total distance covered. Step 1: Calculate distance during acceleration. - Initial velocity $u=0$. - Acceleration $a=1\ \text{m/s}^2$. - Time $t=30\ \text{s}$. - Final velocity $v = u + at = 0 + 1 \times 30 = 30\ \text{m/s}$. - Distance $s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 1 \times 30^2 = 450\ \text{m}$. Step 2: Calculate distance during constant speed. - Speed $v=30\ \text{m/s}$. - Time $t=2\ \text{min} = 120\ \text{s}$. - Distance $s_2 = vt = 30 \times 120 = 3600\ \text{m}$. Step 3: Calculate distance during deceleration. - Initial velocity $u=30\ \text{m/s}$. - Time $t=15\ \text{s}$. - Final velocity $v=0$. - Deceleration $a = \frac{v-u}{t} = \frac{0-30}{15} = -2\ \text{m/s}^2$. - Distance $s_3 = ut + \frac{1}{2}at^2 = 30 \times 15 + \frac{1}{2} \times (-2) \times 15^2 = 450 - 225 = 225\ \text{m}$. Step 4: Total distance $s = s_1 + s_2 + s_3 = 450 + 3600 + 225 = 4275\ \text{m}$. --- 2. **Problem:** A man jogs halfway to his destination at 8 m/s and walks the rest at 6 m/s. Find his average speed for the entire trip. Step 1: Let total distance be $D$. - Jogging distance $= \frac{D}{2}$ at $8\ \text{m/s}$. - Walking distance $= \frac{D}{2}$ at $6\ \text{m/s}$. Step 2: Calculate time for each part. - Jogging time $t_1 = \frac{D/2}{8} = \frac{D}{16}$. - Walking time $t_2 = \frac{D/2}{6} = \frac{D}{12}$. Step 3: Total time $t = t_1 + t_2 = \frac{D}{16} + \frac{D}{12} = D \left(\frac{3}{48} + \frac{4}{48}\right) = \frac{7D}{48}$. Step 4: Average speed $v_{avg} = \frac{\text{total distance}}{\text{total time}} = \frac{D}{\frac{7D}{48}} = \frac{48}{7} \approx 6.86\ \text{m/s}$. --- 3. **Problem:** A bus travels 400 m between two stops. It starts from rest, accelerates at $1.5\ \text{m/s}^2$ until it reaches $9\ \text{m/s}$, continues at this speed, then decelerates at $2\ \text{m/s}^2$ to stop. Find total time. Step 1: Calculate time and distance during acceleration. - Initial velocity $u=0$. - Final velocity $v=9$. - Acceleration $a=1.5$. - Time $t_1 = \frac{v-u}{a} = \frac{9}{1.5} = 6\ \text{s}$. - Distance $s_1 = ut_1 + \frac{1}{2}at_1^2 = 0 + \frac{1}{2} \times 1.5 \times 6^2 = 27\ \text{m}$. Step 2: Calculate time and distance during deceleration. - Initial velocity $u=9$. - Final velocity $v=0$. - Deceleration $a=-2$. - Time $t_3 = \frac{v-u}{a} = \frac{0-9}{-2} = 4.5\ \text{s}$. - Distance $s_3 = ut_3 + \frac{1}{2}at_3^2 = 9 \times 4.5 + \frac{1}{2} \times (-2) \times 4.5^2 = 40.5 - 20.25 = 20.25\ \text{m}$. Step 3: Calculate distance at constant speed. - Total distance $400$ m. - Distance at constant speed $s_2 = 400 - s_1 - s_3 = 400 - 27 - 20.25 = 352.75\ \text{m}$. - Speed $v=9$. - Time $t_2 = \frac{s_2}{v} = \frac{352.75}{9} \approx 39.19\ \text{s}$. Step 4: Total time $t = t_1 + t_2 + t_3 = 6 + 39.19 + 4.5 = 49.69\ \text{s}$. --- 4. **Problem:** A boy throws a ball vertically downward at $10\ \text{m/s}$ from a 20 m high building. (a) Find time to reach ground. (b) Find speed on impact. Step 1: Use equation for displacement $s=20$ m downward, initial velocity $u=10$ m/s downward, acceleration $a=9.8$ m/s² downward. Step 2: Use formula: $$s = ut + \frac{1}{2}at^2$$ $$20 = 10t + 4.9t^2$$ Rearranged: $$4.9t^2 + 10t - 20 = 0$$ Step 3: Solve quadratic: $$t = \frac{-10 \pm \sqrt{10^2 - 4 \times 4.9 \times (-20)}}{2 \times 4.9} = \frac{-10 \pm \sqrt{100 + 392}}{9.8} = \frac{-10 \pm \sqrt{492}}{9.8}$$ Step 4: Calculate positive root: $$t = \frac{-10 + 22.18}{9.8} = \frac{12.18}{9.8} \approx 1.24\ \text{s}$$ Step 5: Calculate final velocity: $$v = u + at = 10 + 9.8 \times 1.24 = 10 + 12.15 = 22.15\ \text{m/s}$$ --- 5. **Problem:** A man drives 30 min at 25 m/s, stops 10 min, then drives 20 min at 36 m/s. Find (a) total distance, (b) average speed. Step 1: Convert times to seconds. - 30 min = 1800 s. - 10 min stop (no distance). - 20 min = 1200 s. Step 2: Calculate distances. - Distance 1: $d_1 = 25 \times 1800 = 45000\ \text{m}$. - Distance 2: $d_2 = 36 \times 1200 = 43200\ \text{m}$. - Total distance $d = 45000 + 43200 = 88200\ \text{m}$. Step 3: Calculate total time including stop. - Total time $t = 1800 + 600 + 1200 = 3600\ \text{s}$. Step 4: Calculate average speed. $$v_{avg} = \frac{\text{total distance}}{\text{total time}} = \frac{88200}{3600} = 24.5\ \text{m/s}$$