A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 7.8 m/s relative to the ground. When the train moves at a constant velocity, the raindrops make an angle of 56° when they move past the window. How fast is the train moving?

To find the speed of the train, we can use trigonometry. Here's how to approach the problem:

Step 1: Assign variables:
Let's call the speed of the train "v" (what we are trying to find), and the speed of the raindrops relative to the ground "r" (given as 7.8 m/s). The angle at which the raindrops move past the window is given as 56°.

Step 2: Analyze the situation:
When the train is stationary, the raindrops fall vertically down. However, when the train moves at a constant velocity, the raindrops appear to move at an angle past the window due to the train's motion.

Step 3: Use trigonometry:
We can use trigonometric functions to relate the known variables and solve for the unknown. In this case, we use the tangent function since we have the opposite and adjacent sides of a right-angled triangle. The tangent of an angle is equal to the opposite side divided by the adjacent side.

The equation for tangent is: tan(θ) = opposite/adjacent

In our case, the opposite side is the speed of the raindrops relative to the ground (r) and the adjacent side is the speed of the train (v).

Step 4: Calculate the speed of the train:
Using the equation from Step 3, we can substitute the given values:

tan(56°) = r / v

Plugging in the values:
tan(56°) = 7.8 m/s / v

Now, we solve for v:
v = 7.8 m/s / tan(56°)

Using a calculator, we can find the value of tan(56°), which is approximately 1.4966.

Finally, divide 7.8 m/s by 1.4966:
v = 7.8 m/s / 1.4966 ≈ 5.21 m/s

So, the train is moving at a speed of approximately 5.21 m/s.

To determine the speed of the train, we can use trigonometry. Let's assume the speed of the train is denoted as v_train.

When the train is stationary, the raindrops fall vertically down, which means their direction of motion is parallel to the ground. So, the angle between the raindrops and the window is 0 degrees.

When the train is moving, the raindrops appear to be traveling at an angle of 56 degrees when they move past the window. This means that the raindrops have a horizontal component of motion due to the train's velocity.

We can use the concept of vector components to solve this problem. The vertical component of the raindrop's velocity is still 7.8 m/s, but it now also has a horizontal component due to the train's velocity.

Let's denote the horizontal component of the raindrop's velocity as v_horizontal.

Using trigonometry, we can write:

cos(56°) = v_horizontal / 7.8 m/s

Rearranging the equation to solve for v_horizontal:

v_horizontal = 7.8 m/s * cos(56°)

Now, the horizontal component of the raindrop's velocity is also equal to the train's velocity:

v_train = v_horizontal

Plugging in the values:

v_train = 7.8 m/s * cos(56°)

Calculating the value:

v_train ≈ 7.8 m/s * 0.5592

v_train ≈ 4.354 m/s

Therefore, the speed of the train is approximately 4.354 m/s.

tan56=v/7.8