Let

*v* = original speed in mi/hr

For 1 hr voyage, distance =

*v* mi

1 hour of full speed + 1 hour of full stop + time to destination at reduced speed = required time + hours delayed

$1 + 1 + \dfrac{500 - v}{0.75v} = \dfrac{500}{v} + 3.75$

$v = \frac{2000}{37} ~ \text{mph}$

Let *x* = distance from the destination where the ship should have been stopped in order to arrive 1.25 hours sooner.

time at full speed + 1 hour of full stop + time to destination at reduced speed = required time + hours delayed - time sooner

$\dfrac{500 - x}{\frac{2000}{37}} + 1 + \dfrac{x}{0.75(\frac{2000}{37})} = \dfrac{500}{\frac{2000}{37}} + 3.75 - 1.25$

$x = 243.24 ~ \text{mi}$ *answer*