A hobo hops a freight train traveling at 2.5 m/s due east. The man boards the train from the rear of the boxcar and walks from the southwest corner to the north east corner of the car at a rate of 5.0 m/s. The boxcar is 15 meters long and 3 meters wide. What is the velocity of the man relative to the train tracks?

Since it isn't specified, I have to assume that when he's walking inside, along the diagonal of the car, at 5m/s, that speed is relative to the floor of the car.  The reason I have to assume that is because I think it makes the problem easier.  I could be wrong.  And it's still troubling, because 5 m/s is a hefty 11.18 mph, which is a pretty energetic walk.  (In fact, it's a 5min 22sec mile, which I for one never accomplished, even when I was running.)  But so be it.-- The length of the car's diagonal is √(3² + 15²) = √(9 + 225) = √334 -- The angle of his walk along the diagonal is the angle whose tangent is 3/15.-- His velocity consists of the components [ 5 cos(angle) east ] and [ 5 sin(angle) north ].That's [ 5 x 15/√334 east ] and  [ 5 x 3/√334 ] north .The train's motion adds to the easterly component of his velocity, and that becomes [ (2.5) + (5 x 15/√334) ] .  The train's motion hasno effect on the northerly component of his velocity.So now we're ready to put the components together and find his velocityrelative to the tracks.  I think it'll be easier to go ahead and get the numericalvalue of each component, and then combine them. Easterly component:  (2.5) + (5 x 15/√334)  = 6.6038 m/sNortherly component: 5 x 3/√334 = 0.8208 m/sJust before finding the magnitude, we note that the direction of his velocityis (the angle whose tangent is 0.8208/6.6038) north of east. That's about 7.085 degrees north of east ...the compass bearing of 82.92 degrees. Now for the magnitude.  It's the square root of the sum of the squares ofthe easterly component and the northerly component.√ (6.6038² + 0.8208²) = √44.2841 = 6.655 m/s(All numbers are rounded.)That's my story, and I'm sticking with it.