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Two students walk in the same direction along a straight path, at a constant speedâone 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?

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boffeemadrid

Answer:

456.143684211 seconds

564.3 m

Explanation:

s = Distance

v = Velocity

Time is given by

[tex]t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{780}{0.9}\\\Rightarrow t=866.67\ s[/tex]

[tex]t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{780}{1.9}\\\Rightarrow t=410.526315789\ s[/tex]

Difference in time = 866.67-410.526315789 = 456.143684211 seconds

According to the question

[tex]\dfrac{x}{0.9}-\dfrac{x}{1.9}=5.5\times 60\\\Rightarrow x(\dfrac{1}{0.9}-\dfrac{1}{1.9})=330\\\Rightarrow x=\dfrac{330}{\dfrac{1}{0.9}-\dfrac{1}{1.9}}\\\Rightarrow x=564.3\ m[/tex]

The students would have to walk 564.3 m