Let's assume a few things first.<br /><br />First: We will be on an earth-sized planet orbiting a sol-sized sun at 1 AU. I do this because I already know some numbers for a system like that.<br /><br />Second: The planet is tidally locked, so there is NO apparent rotation at all -- the sun stays in the same place in the sky if you stay in the same place.<br /><br />Third: The planet has its equator line up with the ecliptic (there is no tilt of 22 degrees, like on earth).<br /><br />Fourth: You are on the equator, at sea level, and there are no elevation changes in the way you will be walking.<br /><br />Fifth: I'll add one more thing: There is no atmosphere to cause bending of sunrays. So you can't see the sun "over the horizon".<br /><br />Okay. Let's say you are starting at a spot on the equator where the sun is just below the horizon. The sun has an angular diameter of about 0.5 degrees. So to make it rise to the point where you are seeing the whole thing, you need to walk 0.5 degrees around the planet. This is 1/720th of the way around. Since earth has a circumference of about 25000 miles at the equator, this means you would have to walk about 34.7 miles to cause the sun to come up.<br /><br />If the planet is rotating on you, the job gets harder. Basically, it would have to be rotating slower than you can walk or run.<br /><br />On earth, to make the sun stand still in a similar scenario, you'd have to be going 25000miles/24hr which is about 1040 miles per hour. An velocity added to or subtracted from that would make the sun rise or set for you.