# I don't understand the approach used in solving this kinematics problem

The problem text is as follows: During the first half of a straight path, which is at $\alpha_1=60°$ relative to a reference line, a car is travelling at $v_1=72 km/h$. During the second half of the straight path which is now at $\alpha_2=30°$ relative to the same reference line, the car is traveling at $v_2=36km/h$. What is the average velocity of the car?

So I've used two different approaches to solve this problem and none of them are correct for reasons unknown to me. The first approach was to find the x and y components of both velocities, then using those find the average x and y velocities, and finally using Pythagoras theorem to find the average velocity using average x and y velocity components. This yields 43.56 which is incorrect. Please note that the expression I used for the average velocity of this kind of motion (one velocity for one half of the path, and another for the 2nd half) is $$\frac{2v_1v_2}{v_1+v_2}$$ (I used this to find average x and y velocity components). Although this approach was intuitive to me I think it is wrong because I don't think that just finding the average velocity components and then using them to find the average velocity is the way to go. I tried simply applying the definition equation of average velocity ($\frac{displacement}{time~interval}$) and then it boils down to simply plugging the velocities in the above written equation and it then the average velocity is 48 km/h. But this is incorrect as well. The correct value of the average velocity is 46.36 km/h. Any clarification would be welcome.

Firstly they tell you that the two bits of the path are the same length ('the first half of the path') and are straight. Call the length $l$. They also give you the angles, $\alpha_1$ and $\alpha_2$ of the two bits of the path to some reference line. This is enough to compute the straight-line distance between the start and finish point in terms of $l$, just using some basic trigonometry.
Now, you know the speeds of the car on the two halves of the path, and you know that $\textit{time}=\textit{distance}/\textit{speed}$. So you know the time the car took to traverse each part of the path, in terms of $l$.