A matatu passes a checkpoint at 9:00 a.m. at 99 km/h. A police car passes the same checkpoint at 9:15 a.m. at 132 km/h. Calculate (i) the relative speed, (ii) the distance between them at 9:15 a.m., a

$$\text{Relative speed} = S_P - S_M$$ $$\text{Relative speed} = 132 \text{ km/h} - 99 \text{ km/h}$$ $$\text{Relative speed} = \boxed{33 \text{ km/h}}$$

$$\text{Distance} = S_M \times \text{Time}$$ $$\text{Distance} = 99 \text{ km/h} \times 0.25 \text{ h}$$ $$\text{Distance} = \boxed{24.75 \text{ km}}$$

and their relative speed to find the time to overtake.

$$\text{Time to overtake} = \frac{\text{Distance between vehicles}}{\text{Relative speed}}$$ $$\text{Time to overtake} = \frac{24.75 \text{ km}}{33 \text{ km/h}}$$ $$\text{Time to overtake} = 0.75 \text{ hours}$$

$$\text{Overtaking time} = 9:15 \text{ a.m.} + 45 \text{ minutes}$$ $$\text{Overtaking time} = \boxed{10:00 \text{ a.m.}}$$

19. (b) Two Passenger Trains

• Initial distance between trains = $240 \text{ m}$
• Train A speed ($S_A$) = $164 \text{ km/h}$
• Train B speed ($S_B$) = $88 \text{ km/h}$
• Train A length ($L_A$) = $150 \text{ m}$
• Train B length ($L_B$) = $100 \text{ m}$

Since they are approaching each other, their speeds add up.

$$\text{Relative speed} = S_A + S_B$$ $$\text{Relative speed} = 164 \text{ km/h} + 88 \text{ km/h} = 252 \text{ km/h}$$

$$\text{Relative speed} = 252 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}$$ $$\text{Relative speed} = 252 \times \frac{5}{18} \text{ m/s} = 70 \text{ m/s}$$

This includes the initial distance between them and their combined lengths.

$$\text{Total distance} = \text{Initial distance} + L_A + L_B$$ $$\text{Total distance} = 240 \text{ m} + 150 \text{ m} + 100 \text{ m} = 490 \text{ m}$$

$$\text{Time} = \frac{\text{Total distance}}{\text{Relative speed}}$$ $$\text{Time} = \frac{490 \text{ m}}{70 \text{ m/s}}$$ $$\text{Time} = \boxed{7 \text{ seconds}}$$