Start by defining your axes and variables. Let East be $+x$, North be $+y$, measured in meters with time in seconds. If A starts at the origin, B starts at $(350,0)$ As A is riding North, his location at time $t$ is $(0,5t)$ What is B's location after he starts riding? Now find the distance $d$ as a function of time. You are then asked for $\frac {dd}{dt}$ at 25 minutes.
Notice, we have $$T=Ce^{kt}+T_s$$
Setting $T=T_0$ (initial temperature of cold drink) at $t=0$ then we have $$T_0=Ce^{0}+T_s\iff C=T_0-T_s$$
Now, we have $$T=(T_0-T_s)e^{kt}+T_s$$ $$e^{kt}=\frac{T-T_s}{T_0-T_s}$$
$$t=\frac{1}{k}\ln\left(\frac{T-T_s}{T_0-T_s}\right)\tag 1$$
Given condition: Temperature rise of cold drink from $\color{red}{T_0=5^\circ \ C\to T=10^\circ \ C}$ in time $t=25\ minutes$ & $\color{red}{T_s=20^\circ \ C}$ substituting values in (1) we get $$25=\frac{1}{k}\ln\left(\frac{10-20}{5-20}\right)$$
$$k=\frac{1}{25}\ln\left(\frac{2}{3}\right)$$
a) Temperature $\color{red}{T}$ after $t=40 \ minutes$, setting corresponding values in (1), we get
$$40=\frac{1}{\frac{1}{25}\ln\left(\frac{2}{3}\right)}\ln\left(\frac{T-20}{5-20}\right)$$ $$\ln\left(\frac{20-T}{15}\right)=\frac{40}{25}\ln\left(\frac{2}{3}\right)$$ $$\frac{20-T}{15}=\left(\frac{2}{3}\right)^{8/5}$$
$$\color{red}{T}=20-15\left(\frac{2}{3}\right)^{8/5}\approx \color{blue}{12.15^\circ\ C}$$
b) Time $\color{red}{t}$ when the temperature is $\color{red}{T=15^\circ C}$, setting corresponding values in (1), we get
$$t=\frac{1}{\frac{1}{25}\ln\left(\frac{2}{3}\right)}\ln\left(\frac{15-20}{5-20}\right)$$
$$=\frac{25 \ln\left(\frac{1}{3}\right)}{\ln\left(\frac{2}{3}\right)}$$
$$\color{red}{t}\approx \color{blue}{67.74\ minutes}$$
Best Answer
The method is basically sound with one error. In your figure, you have a right triangle with leg lengths $500$ and $4800+4500$. The length of the hypotenuse of that triangle is $\sqrt{(4800+4500)^2 + 500^2}$, not $\sqrt{4800^2 + 4500^2 + 500^2}$.