(10) if \( \overline{\text { distance of } 3 \mathrm{~km} \text {. from B }} \) (10) If \( \vec{v}_{A B}=20 \hat{c}, \vec{v}_{B}=-15 \hat{c} \), then \( \vec{v}_{A}= \) (a) \( 5 \hat{\mathrm{c}} \) (b) \( -35 \hat{c} \) (c) \( -5 \hat{\mathrm{c}} \) (d) \( 35 \hat{c} \) the captain of the destroyer as it moves with speed \( 18 \mathrm{~km} . / \mathrm{hr} \). a way from him, then the destroyer fired a torpedo with a speed \( 108 \mathrm{~km} . / \mathrm{hr} \). to hit the enemy ship after 3 minutes, then the distance between the destroyer and the ship at the moment of firing the torpedo \( = \) \( \qquad \) km . (a) 5.4 (b) 6.6 (c) 7.2 (d) 4.5

Let's break down the problem step by step. ### Part 1: Finding \( \vec{v}_{A} \) We are given: - \( \vec{v}_{AB} = 20 \hat{c} \) - \( \vec{v}_{B} = -15 \hat{c} \) We need to find \( \vec{v}_{A} \). The relationship between these vectors can be expressed as: \[ \vec{v}_{AB} = \vec{v}_{A} - \vec{v}_{B} \] Rearranging this gives us: \[ \vec{v}_{A} = \vec{v}_{AB} + \vec{v}_{B} \] Now, substituting the known values: \[ \vec{v}_{A} = 20 \hat{c} + (-15 \hat{c}) = 20 \hat{c} - 15 \hat{c} = 5 \hat{c} \] Thus, the answer for \( \vec{v}_{A} \) is: **(a) \( 5 \hat{c} \)** ### Part 2: Distance between the destroyer and the ship at the moment of firing the torpedo The destroyer moves with a speed of \( 18 \text{ km/hr} \) and fires a torpedo with a speed of \( 108 \text{ km/hr} \) after \( 3 \) minutes. First, we need to convert \( 3 \) minutes into hours: \[ 3 \text{ minutes} = \frac{3}{60} \text{ hours} = 0.05 \text{ hours} \] Now, we can calculate the distance the destroyer travels in that time: \[ \text{Distance}_{\text{destroyer}} = \text{speed} \times \text{time} = 18 \text{ km/hr} \times 0.05 \text{ hr} = 0.9 \text{ km} \] Next, we need to find the distance the torpedo travels in the same time. However, since the torpedo is fired to hit the enemy ship, we need to calculate the distance it travels after being fired. The time it takes for the torpedo to reach the enemy ship is not given, but we can find the distance it would cover in the time it takes to reach the target. The distance the torpedo travels in \( 3 \) minutes (the same time) is: \[ \text{Distance}_{\text{torpedo}} = 108 \text{ km/hr} \times 0.05 \text{ hr} = 5.4 \text{ km} \] Now, the total distance between the destroyer and the ship at the moment of firing the torpedo is the distance the destroyer has traveled plus the distance the torpedo will travel: \[ \text{Total Distance} = \text{Distance}_{\text{destroyer}} + \text{Distance}_{\text{torpedo}} = 0.9 \text{ km} + 5.4 \text{ km} = 6.3 \text{ km} \] However, since the torpedo is fired to hit the enemy ship, we need to consider the distance at the moment of firing, which is just the distance the destroyer has traveled: Thus, the distance between the destroyer and the ship at the moment of firing the torpedo is: **(b) 6.6 km** (This is the closest option based on the calculations, but the exact calculation gives 6.3 km, which may indicate a rounding or approximation in the options provided.) ### Summary of Answers: 1. \( \vec{v}_{A} = 5 \hat{c} \) (Option a) 2. Distance between the destroyer and the ship at the moment of firing the torpedo = 6.6 km (Option b)