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a passenger train leaves a railway station at 12 p.m and travels 3 hours on the Northbound tracks. A second passenger train leaves the same railway station at 12:45 p.m. and travels 2 hours and 15 minutes on the southbound tracks at a speed of 20 kilometers per hour faster than the first train. At 3:00 pm the two trains were 360 kilometers apart. How fast was each train traveling? include units in answer​

Sagot :

retirEEd

Answer:

Step-by-step explanation:

Train A is northbound at     S km/hr    S = speed            and travels for 3 hours

S is the speed of Train A

Train B is southbound at (S + 20) km/hr    

Train B starts later 45 min and travels for 2 and 1/4 hours    so 3/4 hour plus 2 1/4 hour equals 3 hours

After 3 hours since Train A left the station Train A and Train B are 360 km apart

    A + B = 360 km            Equation 1

    A = S km/hr x 3 hr

    B = (S +20) km/hr x (2 1/4) hr

Plug A and B into the first equation

   3S (km/hr) + (2 1/4)(S + 20)  (km/hr)= 360 km

    3S + (9/4)(S + 20) = 360      i dropped the units to make easier to read

    3S + 9S/4 + 180S/4 = 360   I just multiplied 9/4 time the S +20

    12S + 9S + 180 = 1440        multiply both sides by 4

     12S + 9S = 1440 - 180       subtract 180 from both sides

     21S = 1260                       collecting terms

      S = 60 km/hr    S = Speed of Train A  equal 60 km/hr I added the units back into the answer

      S + 20 is the speed of

     Train B  = 80 km/hr

Extra credit

now using the same train speeds, if Train B was now headed NORTH, a what time would it crash into Train A?

S km/hr x T hr = (S+20) km/hr x (T-3/4) hr     solve for T      T = time in hr

When is Train A equal the same location as Train B

Trani A = Train B     don't forget the start time differential

ST = (S + 20)(T-3/4)         plug in train a and b speeds

60T = 80(T-3/4)

60T = 80T - 60

60 = 80T - 60T

60 = 20T

3 = T

after 3 hrs Train B will catch Train A

     

   

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