Use the formulae provided to calculate the following. Use appropriate units for your answers and round off to one decimal place. a) The average speed of a train that travels in \( 4 \frac{1}{2} \) minutes between two stations which are \( 5,2 \mathrm{~km} \) from each other. b) The speed of light in a vacuum is about 300000 km per second. Calculate in minutes how long it takes for sunlight to reach: i) Earth, which is about 150000000 km from the sun ii) Mars, which is about 230000000 km from the sun. c) A javelin thrower at the Olympic Games launches a javelin at a speed of \( 83 \mathrm{~km} / \mathrm{h} \). How far did the Olympian throw the javelin if it remained above ground for four seconds?

Sure, let's solve each part step by step. --- ### **a) Average Speed of the Train** **Given:** - **Distance (d)** = 5.2 km - **Time (t)** = \(4 \frac{1}{2}\) minutes = 4.5 minutes **Average Speed (v)** is calculated using the formula: \[ v = \frac{d}{t} \] \[ v = \frac{5.2 \text{ km}}{4.5 \text{ minutes}} \approx 1.1556 \text{ km/min} \] To express this speed in **km/h**: \[ 1.1556 \text{ km/min} \times 60 \text{ minutes/hour} \approx 69.3 \text{ km/h} \] **Answer:** \[ \text{Average speed} = 69.3 \text{ km/h} \] --- ### **b) Time for Sunlight to Reach Earth and Mars** **Given:** - **Speed of light (c)** = 300,000 km/s #### **i) Time to Reach Earth** - **Distance to Earth (d₁)** = 150,000,000 km \[ \text{Time (t₁)} = \frac{d₁}{c} = \frac{150{,}000{,}000 \text{ km}}{300{,}000 \text{ km/s}} = 500 \text{ seconds} \] Convert seconds to minutes: \[ 500 \text{ s} \div 60 \approx 8.3 \text{ minutes} \] #### **ii) Time to Reach Mars** - **Distance to Mars (d₂)** = 230,000,000 km \[ \text{Time (t₂)} = \frac{d₂}{c} = \frac{230{,}000{,}000 \text{ km}}{300{,}000 \text{ km/s}} \approx 766.6667 \text{ seconds} \] Convert seconds to minutes: \[ 766.6667 \text{ s} \div 60 \approx 12.8 \text{ minutes} \] **Answers:** - **i) Earth:** 8.3 minutes - **ii) Mars:** 12.8 minutes --- ### **c) Distance Thrown by the Javelin** **Given:** - **Speed (v)** = 83 km/h - **Time (t)** = 4 seconds First, convert the speed from **km/h** to **m/s**: \[ 83 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ hour}}{3600 \text{ s}} = \frac{83 \times 1000}{3600} \approx 23.0556 \text{ m/s} \] Now, calculate the distance using: \[ \text{Distance (d)} = v \times t = 23.0556 \text{ m/s} \times 4 \text{ s} \approx 92.2 \text{ meters} \] **Answer:** \[ \text{Distance thrown} = 92.2 \text{ meters} \] --- ### **Summary of Answers** **a)** 69.3 km/h **b)**  i) 8.3 minutes  ii) 12.8 minutes **c)** 92.2 meters