You are driving south on a highway at 24.0 m/s (approximately 54 mi/h) in a snowstorm. When you last stopped, you noticed that the snow was coming down vertically, but it is passing the windows of the moving car at an angle of 26.6° to the horizontal. Calculate the speed of the snowflakes relative to the car.
Calculate the speed of the snowflakes relative to the ground.
With what speed do the supplies land in the latter case?
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To calculate the speed of the snowflakes relative to the car, we can use trigonometry. Let's denote the speed of the snowflakes relative to the car as V_car. We know the car's speed (V_car) and the angle at which the snowflakes are passing the windows (26.6°).
Using trigonometry, we can break down the velocity vector of the snowflakes into horizontal and vertical components. The vertical component is V_snow * sin(theta), where V_snow is the speed of the snowflakes relative to the ground and theta is the angle the snowflakes make with the horizontal.
Since we are given the angle the snowflakes make with the horizontal, and we want to find the horizontal component of the velocity, we can use cosine, which gives us V_snow * cos(theta).
The speed of the snowflakes relative to the car (V_car) is equal to the horizontal component of the snowflake's velocity. Therefore, we can write:
V_car = V_snow * cos(theta)
Given that V_car = 24.0 m/s and theta = 26.6°, we can rearrange the equation to solve for V_snow:
V_snow = V_car / cos(theta)
Now, we can substitute the given values and calculate the speed of the snowflakes relative to the car:
V_snow = 24.0 m/s / cos(26.6°)
V_snow ≈ 26.4 m/s
So, the speed of the snowflakes relative to the car is approximately 26.4 m/s.
To calculate the speed of the snowflakes relative to the ground, we can use the Pythagorean theorem since the velocity vector of the snowflakes makes a right triangle with the vertical and horizontal components.
Using the Pythagorean theorem, we have:
V_snow^2 = (V_snow relative to ground)^2 + V_car^2
Solving for V_snow relative to ground:
(V_snow relative to ground)^2 = V_snow^2 - V_car^2
(V_snow relative to ground) = √(V_snow^2 - V_car^2)
Substituting the given values:
(V_snow relative to ground) = √(26.4 m/s)^2 - (24.0 m/s)^2)
(V_snow relative to ground) ≈ √(696.96 m^2/s^2 - 576 m^2/s^2)
(V_snow relative to ground) ≈ √120.96 m^2/s^2
(V_snow relative to ground) ≈ 11.0 m/s
So, the speed of the snowflakes relative to the ground is approximately 11.0 m/s.
Finally, to answer the last question, we need more information about the supplies. Are they being dropped from the car or thrown from the car? Additionally, we need to know whether any forces act on the supplies, such as air resistance or the effects of wind. These factors will determine the speed with which the supplies land.