Two students walk in the same direction along a straight path, at a constant speed - one at .90 m/s and the other at 1.90 m/s. a.) assuming that they start at the same point ant the same time, how much sooner does the faster 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?

a)The second student arrive 7 minutes and 36 seconds sooner

b)They have to walk 598,51 m

Step-by-step explanation:

Speed= Distance/Time

We are looking for time now

Time = Distance/Speed

First student

Time = 780 m/0.90 m/s=866,66 seconds

Second student

Time = 780 m/1.90 m/s=410,52 seconds

Difference=866,66-410,52= 456,14 seconds

In minutes

Difference= 456,14 seconds /60=7 minutes 36 seconds

Distance=Speed x Time

Speed¹ x Time¹ =Speed² x Time²

Time¹= Time² + 5.50 min

5.50 min =350 seconds

Time²  =Speed¹ x Time¹ /Speed²

Time²  =Speed¹ x (Time² + 350 seconds  ) /Speed²

Time²  =0.9 x (Time² + 350 seconds  ) /1.9

Time²  =0,4737 x (Time² + 350 seconds  )

Time²  =0,4737 Time² + 165,79 seconds

Time²  - 0,4737 Time² = 165,79 seconds

0,5263 Time² = 165,79 seconds

Time² = 165,79 seconds/0,5263 =315,01 seconds

Distance=1,9 x 315,01 seconds = 598,51 m

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is should be y = 5x

step-by-step explanation:

correct me if im wrong