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Genrish500 [490]
3 years ago
7

A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking 2.50 s. Then security

agents appear, and the man runs as fast as he can back along the sidewalk to his starting point, taking 13.0 s. What is the ratio of the man's running speed to the sidewalk's speed?
Physics
2 answers:
Zigmanuir [339]3 years ago
6 0
13:9 because he starts back then runs slower
snow_tiger [21]3 years ago
6 0

Answer:

The ratio of the man's running speed to the sidewalk's speed: 5: 26.

Explanation:

Let the distance between the sidewalk from one end to the other be x

Distance covered by the man = x

Time taken by the person to cover x distance = 2.50 seconds

Speed of the man while running to other end point = S

S=\frac{x}{2.50 s}..[1]

Distance covered by the man by running back to his starting point after agents appears = x

Time taken by the person to cover x distance after agents appears = 13.0 seconds

Speed of the man while running back to starting point after agents appears = S'

S'=\frac{x}{13.0 s}...[2]

The ratio of the man's running speed to the sidewalk's speed:

[2] ÷ [1]

\frac{S'}{S}=\frac{\frac{x}{13.0 s}}{\frac{x}{2.50 s}}=\frac{5}{26}

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This problem is pure trigonometry. Assuming you know trig, there are only a couple of steps to solving this problem. First, split the velocity into components; recall that any vector not directed along an axis has x and y components. Then, remember that sinΘ = opposite/hypotenuse. Applying this to your scenario, you get sin60° = vy/15. Multiplying this out gives you vy=15sin60. Put this into a calculator (make sure it's set to degree mode because the angle in this problem is in degrees) and you should get 12.99, which you can round up to 13 m/s. This is the velocity in the y-direction.

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By the way, a quick way to find the components of a vector, whether it's velocity, force, or whatever else, is to use these functions. Generally, if the vector points somewhere that's not along an axis, you can use this rule. The x-component of the vector is equal to hypotenuse*cosΘ and the y-component of the vector is equal to hypotenuse*sinΘ.

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