The above question is also aimed at Catastrophe. Thank you for providing the maths involved, unfortunately the subject is not my best of subjects!
It is approximately 36 million miles to Mars at closest approach, which happens only once every twenty-six months. For your question, you need not the distance to Mars, but the distance half way to Mars. That's the distance you need to accelerate before you begin your deceleration. You specified ten meters per second That is a reasonable approximation to the 9.8 meters per second squared we experience on Earth.
So half of 36 Million Miles is 18 Million Miles. At 1.6 Kilometers per mile that is 29 Million Kilometers more or less. But we are dealing with meters, not Kilometers, so multiply the 29 Million by a thousand, so it's 29 Billion Meters to Mars at it's best. Worst is around seven times that.
The formula for time at a constant acceleration over some distance is T = square root of ( X/A) where T is time, X is distance and A is the constant acceleration. This gives a time of approximately T = 53,851 Seconds. At 3,600 Seconds per hour, that leaves you with a time of approximately fifteen hours. You then need another fifteen hours to 'stop'.
This isn't really accurate however. It assumes that the target and launch point are not moving and that there is no force of gravity overall. None of those are true.
The actual numbers can get quite messy.
Earth is moving, and so is Mars. They are also moving at quite different speeds and velocities, then there is gravity to consider. Early on, you have Earths gravity to consider. Accelerate up at one G from Earths surface and you will simply never move relative to your launch point. You will literally never move. So you have to accelerate more at the beginning.
During your entire trip, you are fighting the Suns gravity also, so you don't actually have that nice 1 G for the actual trip. Gravity falls off as the square of the distance from the object. So the Sun's gravity is going down the entire trip, but not by that much. At Mars, it's approximately half of what it is for the Earth.
For a rocket, it's quite hard to keep up a 1 G acceleration too. The rocket motor produces a thrust that is pretty much constant. But the acceleration is the thrust divided by the weight of the rocket at any given second.
That's rocket, cargo (that's you!) and fuel. Early in the trip, most of the weight is fuel.
For the trip you propose, it's an awful lot of fuel. The Space Shuttle burned roughly ten times the weight of the Shuttle in fuel. That was for roughly five minutes at three G's.
For the trip you propose, it would take more fuel that the United States has for this single trip. 7/8's or more of that would be burned just getting into Earth Orbit too.
This is why we use the boost and coast system to get there. It's a long curve to Mars, not a straight line. For the 1G constant you proposed, it would still be a curve, but barely.
And yes, Catastrophe is quite correct. A round trip would take less than a week. But it would use more fuel that the entire world has available at any one time.
I hope this helps. It's as i said not really accurate. to get real numbers, you have to use partial differential equations and several numbers from Astronomy as well. There are astronomers who do this kind of thing often. They will say that I am quite inaccurate. They will be right too. I never even mentioned the influence of the other planets for instance, and totally left out the difference between speed and velocity.
Velocity is speed in a particular direction. It does no good to match speeds with Mars if you have a different velocity.
Oh, and by the way, if we could travel with the drive you mentioned, we wouldn't be going to Mars. Such a thing would make the propulsion system for a starship. Why visit dry and dusty Mars when it could be Alpha Centuri or Tau Ceti? At 1 G it only takes a year to reach the speed of light, or so close to it that you'd need laboratory instruments to tell the difference. So at your 1 G, you are just two years away from anywhere in the Universe.