Superman flies to a planet 5 billion miles away and returns within 5 seconds. More impressive than that is that this wasn't anywhere near his top speed, meaning he could go much faster if he really wanted to.
Since it's 5 billion miles and back that means Superman must've flown around 10 billion miles to accomplish this feat. If he's flying perfectly he should reach the planet in 2.5 seconds, giving him the other 2.5 to return to Earth. In addition, he should reach top speed halfway there and begin decelerating the rest of the way, meaning Superman should reach his maximum velocity during this scene within 1.25 seconds and 2.5 billion miles of the way there. With that in mind we can do the rest of the math pretty easily.
a = (2d)/(t^2)
- d = 2,500,000,000 miles = 4.02e12 meters
- t = 1.25 seconds
(2*4.02e12)/(1.25^2) = 5.15e12 m/s^2
v = (a)(t)
- a = 5.15e12 m/s^2
- t = 1.25 seconds
(5.15e12)(1.25) = 6.44e12 m/s
Result
- Earth-Two Superman's Acceleration: 5.2e12 m/s^2
- Earth-Two Superman's Peak Speed: 21,482c
Tidbit: If Superman were to use this acceleration over even just one meter he would be moving at nearly Mach 9500, over 7x faster than lightning.