1. Problem: A cyclist accelerates from 15 km/h to 30 km/h over 1.3 km. Find acceleration $k$ in $ms^{-2}$. Step 1: Convert speeds to $ms^{-1}$. $15 \text{ km/h} = \frac{15 \times 1000}{3600} = 4.167 ms^{-1}$ $30 \text{ km/h} = \frac{30 \times 1000}{3600} = 8.333 ms^{-1}$ Step 2: Use the kinematic equation for constant acceleration: $$v^2 = u^2 + 2as$$ where $v=8.333$, $u=4.167$, $s=1300$ m. Step 3: Solve for $a$: $$a = \frac{v^2 - u^2}{2s} = \frac{8.333^2 - 4.167^2}{2 \times 1300} = \frac{69.44 - 17.36}{2600} = \frac{52.08}{2600} = 0.02003 ms^{-2}$$ 2. Problem: A cheetah accelerates from rest to 108 km/h in 25 m. (a) Find acceleration. Step 1: Convert speed: $108 \text{ km/h} = \frac{108 \times 1000}{3600} = 30 ms^{-1}$ Step 2: Use $v^2 = u^2 + 2as$ with $u=0$, $v=30$, $s=25$: $$a = \frac{v^2 - u^2}{2s} = \frac{900}{50} = 18 ms^{-2}$$ (b) Assumption: Acceleration is constant over the 25 m. 3. Problem: Train starts from rest, reaches 30 ms$^{-1}$ in 20 s, then travels at this speed for 1 minute. (a) Find acceleration in first 20 s: $$a = \frac{v - u}{t} = \frac{30 - 0}{20} = 1.5 ms^{-2}$$ (b) Total distance: Distance during acceleration: $$s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 1.5 \times 20^2 = 300 m$$ Distance at constant speed: $$s_2 = vt = 30 \times 60 = 1800 m$$ Total distance: $$s = s_1 + s_2 = 300 + 1800 = 2100 m$$ (c) Assumptions: Acceleration is constant during first 20 s; speed is constant during next 60 s. 4. Problem: Driver at 70 km/h brakes with deceleration 5 ms$^{-2}$, 60 m from accident. Does she stop in time? Step 1: Convert speed: $70 \text{ km/h} = \frac{70 \times 1000}{3600} = 19.44 ms^{-1}$ Step 2: Use $v^2 = u^2 + 2as$ with $v=0$, $u=19.44$, $a=-5$: $$0 = 19.44^2 + 2 \times (-5) \times s$$ $$s = \frac{19.44^2}{2 \times 5} = \frac{378}{10} = 37.8 m$$ Step 3: Since 37.8 m $<$ 60 m, driver stops before accident. Distance to stop before accident: $$60 - 37.8 = 22.2 m$$ 5. Problem: Car at 20 ms$^{-1}$ requires 30 m braking distance. (a) Find deceleration: $$0 = 20^2 + 2as \Rightarrow a = \frac{-400}{2 \times 30} = -6.67 ms^{-2}$$ (b) Time to stop: $$v = u + at \Rightarrow 0 = 20 - 6.67t \Rightarrow t = \frac{20}{6.67} = 3 s$$ 6. Problem: Particle velocity increases from 2 ms$^{-1}$ to 16 ms$^{-1}$ in 10 s. Find acceleration: $$a = \frac{v - u}{t} = \frac{16 - 2}{10} = 1.4 ms^{-2}$$