Question is in description.

Question

A train leaves San Diego at 1:00 PM. A second train leaves the same city in the same direction at 3:00 PM. The second train travels 30 mph faster than the first. If the second train overtakes the first at 6:00 PM, what is the speed of each of the two trains?

Answer 1

Distance equals speed * time, and the distances covered by each train (1 and 2) is equal at 6:00 PM distance1 = distance 2 speed1 * time1 = speed2 * time2 speed1 * time = speed2 * (time - 2) since 1st train leaves 2 hours earlier speed1 = speed2 - 30 Solving the 2 equations... v1*t = v2*(t-2) & v1 = v2 - 30 --> v1 = 45 mph and v2 = 75 mph

Answer 2

Hi Tori, ready for some math?

 

Let's start by virtually inspecting the problem

 

 

 Starting Point                                                    Overtaking Point

 for both A

     |                                                                            |

     |                                                                            |

     |                                                                            |

     |                                                                            |

     |                                                                            |

  Train A (1:00 PM) V_a                                               Train A (6:00 PM)

.......................................................................................................

 

Starting Point                                                     Overtaking Point
for B 
    |                                                                             |
    |                                                                             |
    |                                                                             |
    |                                                                             |
    |                                                                             |
Train B (3:00 PM) V_b                                               Train B/Train A (6:00 PM)
.......................................................................................................

 

We know that train B goes 30mph faster than train A

Vb = Va + 30    -----------(1)

 

 

The times for the trip are  ta = 5hrs (from 1:00 PM to 6:00 PM) and tb = 3 hrs (from 3:00 PM to 6:00 PM)

   

Let x be the distance traveled by both trains until train B overtakes train A equal  distances

 

 Xa = Xb  and let t be the times from starting to overtaking remember X = V*t

 

Va*ta = Vb*tb

 

Va*5 = Vb*3

 

Va  = (3/5)Vb plug this into  (1) above

 

Vb= (3/5)Vb + 30

Vb-(3/5)Vb = 30

[1-3/5]Vb = 30

[2/5]Vb = 30

Vb = (5/2)*30mph

The speed of the second train is 75mph and the speed of the first train is 45mph