Activity 1.1 A man on a train is walking to the back of a train at a speed of \( 2 \mathrm{~m} / \mathrm{s} \) while the train is moving forward at \( 10 \mathrm{~m} / \mathrm{s} \), Calculate: a) The man's velocity relative to the train. b) The train's velocity relative to the ground. c) The man's velocity relative to the ground.

Let's analyze the problem step-by-step using the concepts of relative velocity. ### Given: - **Train's velocity relative to the ground (\( V_{\text{train/ground}} \))**: \( 10 \, \text{m/s} \) forward - **Man's velocity relative to the train (\( V_{\text{man/train}} \))**: \( 2 \, \text{m/s} \) toward the back of the train We'll assume **forward** is the positive direction. --- ### a) **The Man's Velocity Relative to the Train** This part is directly given in the problem. - **Velocity (\( V_{\text{man/train}} \))**: \( 2 \, \text{m/s} \) backward Since backward is opposite to the positive (forward) direction: \[ V_{\text{man/train}} = -2 \, \text{m/s} \] **Answer:** The man's velocity relative to the train is **2  m/s toward the rear** of the train. --- ### b) **The Train's Velocity Relative to the Ground** This information is provided directly in the problem statement. - **Velocity (\( V_{\text{train/ground}} \))**: \( 10 \, \text{m/s} \) forward **Answer:** The train's velocity relative to the ground is **10  m/s forward**. --- ### c) **The Man's Velocity Relative to the Ground** To find the man's velocity relative to the ground (\( V_{\text{man/ground}} \)), we use the principle of relative velocity: \[ V_{\text{man/ground}} = V_{\text{train/ground}} + V_{\text{man/train}} \] Plugging in the known values: \[ V_{\text{man/ground}} = 10 \, \text{m/s} \, (\text{forward}) + (-2 \, \text{m/s}) \, (\text{backward}) \] \[ V_{\text{man/ground}} = 10 \, \text{m/s} - 2 \, \text{m/s} = 8 \, \text{m/s} \, \text{forward} \] **Answer:** The man's velocity relative to the ground is **8  m/s forward**. --- ### **Summary of Answers:** a) **2 m/s toward the rear of the train** b) **10 m/s forward** c) **8 m/s forward**