= I dont know how to work it out :rolleyes:

BTW
It cant be done. Would have to return at an infinite speed.

You are mistaken. As layback40 said, the distance traveled is equal to the area under a velocity/time graph. If you draw a curve that has steep acceleration at the beginning with less acceleration at the end, you will get a distance greater than the distance traveled by a vehicle with constant acceleration.

You can check this by assuming different equations for acceleration as a function of time and then integrating them with respect to time. That's another way of saying the same thing layback40 said.

Bingo !!!

I didnt wish to introduce calculus, thought it best to try & keep it simple.

There was no mention of acceleration curves in the OP. Only a speed and a timeto reach that speed.
But you and others that think the answer given is wrong should cough up your own answer.

His answer is only correct if the vehicle has constant acceleration. In that case, a graph of velocity versus time is a straight line with a slope equal to the acceleration. The distance traveled is the area under the curve (i.e. the integral with respect to time.) The area under the linear curve is equal to the area under a curve that has a constant velocity of 50. Draw it and you will see. Then draw a velocity curve that is steep at the beginning and tapering off at the end. The area under that curve will be greater than the area under the linear curve. That result stands to reason because the car with the quicker acceleration at the beginning will spend more of its time traveling at high speed, even though it will not continue to gain speed as quickly as the other car will at the end of the trip.That's true only if acceleration is constant.

You know it's quite late to put on the pedantic act.

Calculus is underrated. Even though few people ever use it after getting out of school, it helps give a better understanding of physical processes, IMHO.

Honus,
Its a waste of time trying to explain, your are just being dragged down to .................. & beaten with experience.

I already did. See post #5.

ACTUALLY.....
you simply drive 50 mph on the way back...
because you said ' you wish to average 50 mph for a return journey' ..
thus that stupid funeral you got stuck behind had nothing to do with the return trip... except to move its start time to later than you wanted...

:confused:

layback, is that what you were talking about?

That agrees with the other answers. But you later claim they are wrong.

Don't get distracted by strawman examples.

Not the return journey !!!
You know what I meant. Average for going over & back !!
Not a silly play on words.

BTW the average of 2 numbers is calculated by adding them & then dividing by 2 . Not taking one from the other, that only works when one of the numbers is 0.

I think you have the right idea.

I have asked our fellow member to give some details of his education in another thread ~ no such information is forth coming.