A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 4.15 m/s relative to the ground. When the train moves at a constant velocity, the raindrops make an angle of 25° when they move past the window, as the drawing shows. How fast is the train moving?

V/4.15 = tan 25

Do you see why? Think about it.

Solve for V

Well, it seems the raindrops are playing a little game of "Let's make this person do some math!" with the person on the train. Sneaky raindrops, I must say.

To figure out how fast the train is moving, we need to break out some trigonometry. The raindrops are falling vertically at a speed of 4.15 m/s, relative to the ground. When the train moves, the raindrops make an angle of 25° with respect to the window.

So, let's call the speed of the train "v." Now, we can use some trigonometric magic called sine to relate the vertical speed of the raindrops to the speed of the train.

The vertical speed of the raindrops is equal to the speed of the train multiplied by the sine of the angle between the raindrops and the window. Mathematically, it looks like this:

4.15 m/s = v * sin(25°)

Now, let's solve for the speed of the train, v:

v = 4.15 m/s / sin(25°)

Alright, if you're still with me, let's do some calculator work. *beep boop beep* According to my calculations, the train is moving at approximately 9.55 m/s.

So there you have it! The train is moving at a speed of around 9.55 m/s. Just remember, next time it rains, those mischievous raindrops might be trying to teach you some trigonometry!

To solve this problem, we can use trigonometry.

Let's denote the speed of the train as Vt and the speed of the raindrop relative to the train as Vr. We are given that Vr = 4.15 m/s, and the raindrops make an angle of 25° with the vertical when viewed from the moving train.

We need to find the value of Vt.

We can break down the raindrop's velocity into two components: one perpendicular to the train's motion, Vr⊥, and one parallel to the train's motion, Vr∥. The raindrop's velocity relative to the train can be expressed as follows:

Vr⊥ = Vr * sin(25°)
Vr∥ = Vr * cos(25°)

Now, since the train is moving at a constant velocity, the raindrop's vertical speed relative to the ground must match the train's vertical speed. Therefore, we can write:

Vt = Vr⊥ = Vr * sin(25°)

Plugging in the given value for Vr, we have:

Vt = 4.15 m/s * sin(25°)
Vt ≈ 1.76 m/s

Therefore, the speed of the train is approximately 1.76 m/s.

To find the speed of the train, we can use trigonometry. Let's denote the speed of the train as v.

From the perspective of the person on the train, the raindrops appear to be falling vertically downwards. However, when observing from the ground, the raindrops appear to be falling at an angle due to the train's velocity.

In the drawing, the angle between the vertical direction and the raindrop's path is given as 25°. Hence, we can determine that the vertical component of the raindrop's velocity is v * sin(25°) and the horizontal component is v * cos(25°).

Given that the speed of the raindrops relative to the ground is 4.15 m/s, we can equate the horizontal component of their velocity to this value:

v * cos(25°) = 4.15 m/s

Solving this equation will give us the value of v, which represents the speed of the train.

To calculate v, we rearrange the equation:

v = 4.15 m/s / cos(25°)

Using a calculator, we find that cos(25°) is approximately 0.9063. Therefore:

v = 4.15 m/s / 0.9063 ≈ 4.57 m/s

Hence, the speed of the train is approximately 4.57 m/s.