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antoniya [11.8K]
3 years ago
12

One runner in a relay race has a 1.50 s lead and is running at a constant speed of 3.25 m/s. The runner has 30.0 m to run before

reaching the end of the track. A second runner moves in the same direction as the leader. What constant speed must the second runner maintain in order to catch up to the leader at the end of the race?
Physics
2 answers:
yarga [219]3 years ago
6 0
The second runner must run 3.3m/s. If the leading runner is 1.5 seconds ahead and there are 30m left, the second runner would need to run slightly faster than the lead in order to finish at the same time. To calculate this I did 30/1.5 which gave me 0.05. I added this onto the speed of the lead runner to get 3.3m/s :)
Firlakuza [10]3 years ago
5 0

Answer:

3.88m/s

Explanation:

The first runner has a 1.5s lead, multiply it by the constant speed (1.5 x 3.25) this equates to 4.875 metres.

4.875 metres is the distance the first runner has already covered.

Subtract this from the 30metres left to run, leaves = 25.125 metres.

If you divide 25.125 by 3.25 seconds which is constant speed of the first runner you will get the time in which the first runner finished the race.

25.125 / 3.25 = 7.73 seconds.

If you divide 30metres (because the second runner doesn't have a lead same as the first runner) by 7.73 seconds, you get 3.88.

The second runner must move with a speed of 3.88m/s in order to catch up the leader at the end of the race.

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Answer:

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Explanation:

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Remember that in a circle the area is defined as:

A = \pi r^{2}  (1)

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An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.  

   

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

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A_{mirror1} = \pi (\frac{8.4m}{2})^{2}    

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<em>Key term:</em>

<em>Photons: particles that constitute light. </em>

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