Assume that rain is falling at a speed of 5 meters per second \((\mathrm{m} / \mathrm{s})\) and you are driving in the rain at the same speed, a leisurely \(5 \mathrm{m} / \mathrm{s}(18 \mathrm{km} / \mathrm{h}) .\) Estimate the angle from the vertical at which the rain appears to be falling.

In physics, velocity is a vector quantity. This means it has both a magnitude and a direction.
Understanding this is crucial when analyzing movement because two objects moving at the same speed might not necessarily have the same velocity if they are moving in different directions.
In the exercise about rain and the car, each has a velocity of 5 meters per second \((\frac{m}{s})\), but the rain falls vertically while the car moves horizontally.
To find how the rain appears to someone in the car, you need to think about these two velocity vectors combined.
This combined effect is handled using the concept of relative velocity.
Always remember:

• Speed is scalar, which means it only has magnitude.
• Velocity is vector, meaning it has both magnitude and direction.
• Relative velocity helps to analyze the motion of objects from different reference points.

The Pythagorean theorem is a useful tool in physics, especially when dealing with perpendicular vectors.
It's represented by the formula: \[a^2 + b^2 = c^2\]
In the exercise, the velocities of the rain and the car are perpendicular.
This means, if you draw them as vectors, they form a right-angled triangle.
The car's horizontal speed and the rain's vertical speed become the two shorter sides (legs) of the triangle, and the resultant speed is the hypotenuse.
To find this resultant speed, we apply the Pythagorean theorem as follows:

• The horizontal speed of the car: 5 meters per second
• The vertical speed of the rain: 5 meters per second
• Resultant speed (hypotenuse): \[\sqrt{(5)^2 + (5)^2} = \sqrt{50} = 5\sqrt{2} \]

This combined velocity shows the actual speed and direction of the rain from the perspective of someone in the car.

Trigonometry helps in finding angles between vectors, especially in relative motion situations.
In this case, once we have the resultant velocity of the rain as seen from the car, we can use trigonometry to find the angle of the rain with the vertical.
The function we use here is the tangent, given by: \[\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{horizontal speed}}{\text{vertical speed}} \]
From the problem:

• Horizontal speed of the car: 5 meters per second
• Vertical speed of the rain: 5 meters per second

The equation becomes: \[\tan(\theta) = \frac{5}{5} = 1 \]
Then, we find the angle: \[\theta = \arctan(1) = 45\degree \]
This tells us the rain appears to fall at a 45-degree angle from the vertical to someone driving in the car.
Using trigonometry in this way allows us to fully understand the direction and magnitude of relative motion, making it easier to visualize complex concepts.
In summary, always remember:

• Tangents help find angles between perpendicular vectors.
• Arc functions like \(\arctan\) (inverse tangent) are used to get the angle from the ratio of speeds.