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2 months ago

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Two students walk in the same direction along a straight path, at a constant speed one at 0.90 m/s and the other at 1.90 m/s. a.

Assuming that they start at the same point and the same time, how much sooner does the faster student arrive at a destination 780 m away? b. How far would the students have to walk so that the faster student arrives 5.50 min before the slower student?

1 answer:

Sav [3.1K]2 months ago

8 0

Answer: a) 456.66 s ; b) 564.3 m

Explanation:

The time taken to travel a fixed distance at constant speed is calculated by:

velocity = distance / time, thus time = distance / velocity

For the slower student:

t = 780 m / 0.9 m/s = 866.66 seconds

For the faster student:

t = 780 m / 1.9 m/s = 410.52 seconds

The difference in arrival time is 866.66 - 410.52 = 456.14 seconds.

To find the walking distance where the faster student arrives 5.5 minutes earlier, use:

distance = velocity_slow × time_1

distance = velocity_fast × time_2

The time difference corresponds to 5.5 minutes or 330 seconds.

From these, we determine:

time_1 = 627 seconds

time_2 = 297 seconds

The distance traveled in this case is 564.3 m.

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At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

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