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Finger [1]
3 years ago
8

You and a friend work in buildings five equal-length blocks apart, and you plan to meet for lunch. Your friend strolls leisurely

at 1.2 m/s, while you like a brisker pace of 1.6 m/s. Knowing this, you pick a restaurant between the two buildings at which you and your friend will arrive at the same instant if both of you leave your respective buildings at the same instant. In blocks, how far from the wife’s building is the restaurant?
Physics
1 answer:
Pavel [41]3 years ago
7 0

Answer:

Your friend is 2.143 blocks from the restaurant.

You are 2.857 blocks from the restaurant.

Explanation:

Let t be the time both you and your friend take to walk to the restaurant.

The distance (m) from your building to the restaurant is your walking time t times your speed v1

s_1 = v_1t = 1.6t

Similarly the distance (m) from your friend building to the restaurant:

s_2 = v_2t = 1.2t

Let b be the length (in m) of a block, the total distance of 5 blocks is 5b

s_1 + s_2 = 5b

1.6t + 1.2t = 5b

2.8t = 5b

t = 5b/2.8 = 25b/14

s_2 = 1.2t = 1.2(25b/14) = 2.143b

So your friend are 2.143b meters from the restaurant, since each block is b meters long, 2.143b meters would equals to 2.143b/b = 2.143 blocks. And you are 5 - 2.143 = 2.857 blocks from the restaurant.

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The star with apparent magnitude 2 is more brighter than 7.

To find the answer, we have to know about apparent magnitude.

<h3>What is apparent magnitude?</h3>
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Answer:

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Explanation:

This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period

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indicate that the initial velocity is zero

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in this part the final velocity is zero (v₂ = 0)

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let's calculate

         x₂ = \frac{ 143.666^2 }{2 \  18.2}

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