(b) 6.6 A bicyclist is moving in a straight line with uniform speed, watches a car in the opposite direction it seems to him that is moving towards him with a speed of \( 120 \mathrm{~km} / \mathrm{hr} \). and at the same moment he watches a truck in front of him in his direction it seems to him that is moving with speed 60 km ./hr. away from him , then magnitude of the velocity of the car with respect to the truck \( =\ldots \ldots \ldots . . \mathrm{m} / \mathrm{sec} \) \( \begin{array}{llll}\text { (a) } 180 & \text { (b) } 50 & \text { (c) } 60 & \text { (d) } \frac{50}{3}\end{array} \)

To solve this problem, we need to deduce the speed of the bicyclist, car, and truck using relative speeds. Given: - The car appears to move towards the bicyclist at 120 km/h. - The truck appears to move away from the bicyclist at 60 km/h. Let \( v_b \) be the speed of the bicyclist, \( v_c \) be the speed of the car, and \( v_t \) be the speed of the truck. From the perspective of the bicyclist: 1. The relative speed of the car with respect to the bicyclist is given by: \[ v_c + v_b = 120 \text{ km/h} \] 2. The relative speed of the truck with respect to the bicyclist is: \[ v_t - v_b = 60 \text{ km/h} \] We need to find the magnitude of the car’s velocity with respect to the truck: **Step 1**: Express \( v_c \) and \( v_t \) in terms of \( v_b \): - From the first equation, we get: \[ v_c = 120 - v_b \] - From the second equation, we have: \[ v_t = 60 + v_b \] **Step 2**: Find the relative velocity of the car with respect to the truck: - The relative velocity of the car with respect to the truck is given by: \[ v_{ct} = v_c - v_t \] Substituting the expressions for \( v_c \) and \( v_t \): \[ v_{ct} = (120 - v_b) - (60 + v_b) = 120 - v_b - 60 - v_b = 60 - 2v_b \] **Step 3**: Find \( v_{ct} \) in \( m/s \): To convert \( km/h \) to \( m/s \), we factor in \( \frac{1000}{3600} = \frac{5}{18} \): \[ v_{ct} = \left(60 - 2v_b\right) \times \frac{5}{18} \text{ m/s} \] Next, let's analyze the options: 1. If we assume different possible maximum speeds for \( v_b \) that make \( 60 - 2v_b \) positive, we could see how they play out. For example, if \( v_b = 30 \) km/h, then: \[ v_{ct} = 60 - 60 = 0 \text{ km/h} \Rightarrow \text{ (not valid since needs to be > 0)} \] 2. For \( v_b < 30 \) km/h, it gives valid values that align with the options. Assuming the maximum bicyclist speed we can reasonably equate to \( 60 \text{ km/h}\) speeds the truck (with considerable negative speed impacts). To summarize: Setting \( v_b \) minimally at around \( 0 \text { km/hour}\) gives closer to values of \( 60 \text {km/h}\) maxing out overall vehicle speeds in the equation solving yields \(60\) yielding value \(50\) and ultimately \(120 - \frac{100}{60}\) as valid. Final Determination gives the overall choice: **The magnitude of the velocity of the car concerning the truck is \( \text{d) } \frac{50}{3} \text{ m/s}**