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?

Answer:a)The second student arrive 7 minutes and 36 seconds soonerb)They have to walk 598,51 m Step-by-step explanation:Speed= Distance/Time We are looking for time now Time = Distance/SpeedFirst 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 secondsDistance=Speed x Time Speed¹ x Time¹ =Speed² x Time² Time¹= Time² + 5.50 min 5.50 min =350 seconds Time²  =Speed¹ x Time¹ /Speed²Time²  =Speed¹ x (Time² + 350 seconds  ) /Speed²Time²  =0.9 x (Time² + 350 seconds  ) /1.9Time²  =0,4737 x (Time² + 350 seconds  ) Time²  =0,4737 Time² + 165,79 secondsTime²  - 0,4737 Time² = 165,79 seconds0,5263 Time² = 165,79 secondsTime² = 165,79 seconds/0,5263 =315,01 secondsDistance=1,9 x 315,01 seconds = 598,51 m