A salesman drives from Ajax to Barrington, a distance of 115 mi, at a steady speed. He then increases his speed by 14 mi/h to drive the 158 mi from Barrington to Collins. If the second leg of his trip took 8 min more time than the first leg, how fast was he driving between Ajax and Barrington?

Question

anonymous · Answer

assuming steady rise in speed or instantly went from 14 m/h to  x mi/h?

anonymous · Answer

i get 5.733 but I think I am wrong i'll dbl check

anonymous · Answer

now I get 46 for original speed :)

anonymous · Answer

sounds more reasonable 46 mi/h :)

anonymous · Answer

115=vt
sfr t 
t=115/v

here is where it gets complicated the second leg of trip is
158=(v+14)*(t + 2/15)

158=(v+14)*(115/v + 2/15)
sfv =46

anonymous · Answer

@timo86m Thanks!!

anonymous · Answer

was it right?

anonymous · Answer

Yes!

anonymous · Answer

a cool

I think that means he was going 46 + 14 the second leg of trip that is fast :P