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2 years ago

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Calculate the average speed and average velocity of a complete round-trip in which the outgoing 250 km is covered at 95 km/h fol

lowed by a 1.0-hour lunch break, and the return 250 km is covered at 55 km/h.

1 answer:

ser-zykov [4K]2 years ago

5 0

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From them we will consider speed as the distance traveled per unit of time. Said unit of time will be cleared to find the total time taken to travel the given distance. Later with the calculated average times and distances, we will obtain the average speed.

PART A)

The time taken to travel a distance of 250km with a speed of 95km/h is

$t = \frac{d}{v}$

$t = \frac{250km}{95km/h}$

$t = 2.63h$

Time taken for the lunch is

$t = 1h$

The time taken travel a distance of 250km with a speed of 55km/h

$t = \frac{d}{v}$

$t = \frac{250}{55}$

$t = 4.54h$

The total time taken is

$t = t_{outgoing}+t_{lunch}}t_{return}$

$t = 2.63+1+4.54$

$t = 8.17h$

The average speed is the ratio of total distance and total time

$v = \frac{250+250}{8.17}$

$v = 61.15km/h$

PART B)

As the displacement is zero the average velocity is zero.

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Answer:

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We know that the moon makes a full circumference ( $2\,\pi\,R$ ) in 388800 seconds, therefore its tangential velocity is:

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Answer:

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[1000(1 - 3/1000)]^1/3

1000^1/3 * (1 - 3/1000)^1/3

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Answer:

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