To solve this problem, we can use the student's observations about the changing angles of the rain streaks on the windows and the known acceleration of the car. Let's break down the solution into steps:

1. Determine the car's velocity over time:

   • The car accelerates at $a = 1.3 \, \text{m/s}^2$.
   • The time interval is $\Delta t = 2.0 \, \text{s}$.
   • Initial velocity ($u$) is zero since the car is starting from rest.
   • Final velocity ($v$) can be calculated using the equation $v = u + at$.

2. Calculate the horizontal component of the velocity of the rain at different angles:

   • At $t = 0$, the angle is $\theta_1 = 45^\circ$. The car's velocity is zero, so the horizontal component of the rain's velocity is also zero.
   • At $t = 2.0 \, \text{s}$, the angle changes to $\theta_2 = 60^\circ$. The horizontal component of the velocity of the rain is now equal to the velocity of the car.

3. Determine the speed of the rain:

   • Let $v_r$ be the vertical speed of the rain, which remains constant.

Let's work through the calculations:

1. Car's Velocity After 2 Seconds

The car's final velocity after 2 seconds can be calculated using the formula:

$v = u + at$

Since the car starts from rest, $u = 0$. Thus:

$v = 0 + (1.3 \, \text{m/s}^2)(2.0 \, \text{s})$

$v = 2.6 \, \text{m/s}$

This is the horizontal velocity of the car (and the rain relative to the car) at $t = 2.0 \, \text{s}$.

2. Horizontal and Vertical Components of Rain's Velocity

At $\theta_2 = 60^\circ$, the horizontal component of the rain's velocity relative to the car is equal to the car's velocity, which is $2.6 \, \text{m/s}$.

Let $v_r$ be the vertical speed of the rain. The relationship between the horizontal and vertical components can be represented by:

$\tan(60^\circ) = \frac{v_{\text{horizontal}}}{v_r}$

We know $v_{\text{horizontal}} = 2.6 \, \text{m/s}$, so:

$\tan(60^\circ) = \frac{2.6}{v_r}$

$\sqrt{3} = \frac{2.6}{v_r}$

$v_r = \frac{2.6}{\sqrt{3}}$

3. Calculate the Speed of the Rain

Now, we calculate $v_r$:

$v_r = \frac{2.6}{\sqrt{3}} \, \text{m/s}$

$v_r = \frac{2.6}{1.732} \, \text{m/s}$

$v_r \approx 1.5 \, \text{m/s}$

Thus, the speed of the falling rain is approximately 1.5 m/s.

Would you like more details on any part of the calculation, or do you have any other questions?

Here are some related questions you might find interesting:

1. How would the answer change if the acceleration of the car were different?
2. What assumptions are made in this calculation that might affect accuracy?
3. How can the student's observation method be improved for more precise measurement?
4. How would the observed angles change if the rain were not falling vertically?
5. What is the effect of wind on the rain's apparent speed and direction?
6. How would the speed of the rain be affected if the car was moving uphill or downhill?
7. Can we determine the distance traveled by the car using the given data?
8. How does the angle of the rain streaks relate to the relative velocities of the car and rain?

Tip: When solving kinematics problems, it's crucial to differentiate between relative and absolute velocities and ensure that angles are measured relative to a consistent reference frame.