(a) The TOUR de France is happening on the 26th June 2021 from BREST to MÜR DE BRETAGNE. Mark cycles from Brest to MÜR DE BRETAGNE and Jay cycles to the opposite direction. Knowing that the distance between BREST and MÜR DE BRETAGNE is approximately 131 Km and that both cyclists start cycling towards each other at the same time. Assuming that Mark cycles twice as fast as Jay does. If both cyclists meet one and half hour later, then at what average speed is each one of them traveling? (b) Suppose that a body with \( T_{1} \) temperature is placed in surroundings with temperature \( T_{0} \) different from that of \( T_{1} \). The body will either cool or warm to temperature \( T(t) \) after time \( t \), in minutes, where \[ T(t)=T_{0}+\left(T_{1}-T_{0}\right) e^{-k t} \text {. } \] A cup of coffee with temperature \( 40.56^{\circ} \mathrm{C} \) is placed in a freezer with temperature \( -17.78^{\circ} \mathrm{C} \). After 5 min , the temperature of the coffee is \( 21.11^{\circ} \mathrm{C} \). What will its temperature be after 10 min ?

To solve the problems, we will break them down into parts (a) and (b). ### Part (a) Let \( v_J \) be the speed of Jay, and since Mark cycles twice as fast, his speed \( v_M = 2v_J \). The total distance between Brest and Mür de Bretagne is 131 km. They meet after 1.5 hours (or 1.5 hours = 90 minutes). The distance covered by both cyclists when they meet can be expressed as: \[ \text{Distance} = \text{Speed} \times \text{Time} \] The total distance covered by both cyclists when they meet is: \[ \text{Distance} = v_M \cdot t + v_J \cdot t = (2v_J + v_J) \cdot t = 3v_J \cdot t \] Substituting \( t = 1.5 \) hours: \[ 3v_J \cdot 1.5 = 131 \] This simplifies to: \[ 4.5v_J = 131 \] Now, solving for \( v_J \): \[ v_J = \frac{131}{4.5} \approx 29.11 \text{ km/h} \] Now, we can find \( v_M \): \[ v_M = 2v_J = 2 \cdot 29.11 \approx 58.22 \text{ km/h} \] Thus, the average speeds of the cyclists are: - Jay's speed \( v_J \approx 29.11 \text{ km/h} \) - Mark's speed \( v_M \approx 58.22 \text{ km/h} \) ### Part (b) We are given the temperature function: \[ T(t) = T_0 + (T_1 - T_0)e^{-kt} \] where: - \( T_1 = 40.56^{\circ}C \) (initial temperature of the coffee) - \( T_0 = -17.78^{\circ}C \) (temperature of the freezer) - \( T(5) = 21.11^{\circ}C \) (temperature after 5 minutes) First, we need to find the value of \( k \). We can use the temperature after 5 minutes to find \( k \): \[ T(5) = -17.78 + (40.56 + 17.78)e^{-5k} \] Substituting the known values: \[ 21.11 = -17.78 + 58.34e^{-5k} \] \[ 21.11 + 17.78 = 58.34e^{-5k} \] \[ 38.89 = 58.34e^{-5k} \] Now, solving for \( e^{-5k} \): \[ e^{-5k} = \frac{38.89}{58.34} \approx 0.666 \] Taking the natural logarithm: \[ -5k = \ln(0.666) \] \[ k = -\frac{\ln(0.666)}{5} \approx 0.061 \] Now, we can find the temperature after 10 minutes: \[ T(10) = T_0 + (T_1 - T_0)e^{-10k} \] Substituting the values: \[ T(10) = -17.78 + (40.56 + 17.78)e^{-10 \cdot 0.061} \] Calculating \( e^{-10 \cdot 0.061} \): \[ e^{-0.61} \approx 0.543 \] Now substituting back: \[ T(10) = -17.78 + 58.34 \cdot 0.543 \] Calculating: \[ T(10) = -17.78 + 31.69 \approx 13.91^{\circ}C \] Thus, the temperature of the coffee after 10 minutes will be approximately \( 13.91^{\circ}C \).