Driving time puzzle

[Brown_Eyed_Girl]

@Chessic_Sense

Who said anything about point B being a destination? Unless told otherwise, I’d presume the driver was continuing on to points C, D, and E without stopping.

The OP didn’t specify whether Point B was a destination or a waypoint. However, he also made no mention that there were other points in equation, therefore, I figured it was rather safe to assume that Point B was a destination. You assumed the opposite. shrug

Either way, the fact that you will eventually slow down and come to a stop should be factored in at the final destination (or any waypoint you stop at) because it will add time, not distance, to travel time.* Whether this effect should be factored in to the calculations of Points A to B is for the OP to clarify.

• I don’t have the math skills to come up with a formula to figure out how much more time to add to travel time for slowing to a stop at destination. Is the OP a lead foot or easy on his brakes?

[CJJ]

@Xema

Both maps and GPS receivers take this into account. So, for example, if you ask your GPS receiver to tell you the straight-line distance from NYC to LA, it gives the length of a great circle route across the surface of the earth, not the distance of a tunnel through the earth.

I worked on SW that made distance calculations based on GPS readings. I learned more than i ever wanted to know about the Haversine formula…

[Fubaya]

@Corresponder

Yes, this is the answer I intended. I didn’t think someone would get it so quickly. If the road is hilly, then you actually have to travel more than 60 miles. I noticed the other day that, even with my cruise control on, my GPS would display a slightly lower speed than my speedometer when I was going UP or DOWN a hill. Apparently the GPS only calculates your speed as if you were travelling on a flat terrain, in terms of latitude and longitude; the same is true with a flat map.

It may not have affected anything in this case, but I wouldn’t trust the speedometer too much. They are designed to be accurate to a certain percentage but due to factors like tire pressure, tire wear, and simple safety, they usually err on the side of reading too fast. They’re good enough for what they’re intended for, but they’re definitely not scientific instruments.

[Brown_Eyed_Girl]

@Hampshire

How about the time it takes to get up to 60mph starting from a standstill at point A? You can’t start at point A and instantaneously be going 60mph.

That too. I’m still wondering if the OP was factoring traveling slower than 60 mph at the beginning and end of his trip.

[Mangetout]

It sounds like it’s.comfortably within the margin of error of the equipment, but if this is a puzzle, I’m going to say that the 60mph speed is time-averaged rather than distance-averaged.

[Apex_Rogers]

@Brown_Eyed_Girl

That too. I’m still wondering if the OP was factoring traveling slower than 60 mph at the beginning and end of his trip.

I read the OP as having the car pass both points A and B running at 60 mph, thanks to the very accurate cruise control. The whole premise of the puzzle relies on the assumption that the car is at exactly 60 mph the whole time, which makes the question of why it took longer than exactly 60 minutes interesting. If we include stopping and starting as part of the “course”, the answer would be plainly obvious: You were going less than 60 mph for parts of the trip, so of course it takes longer than 60 minutes to go 60 miles.

The wording of the OP did leave it open to your sort of “lateral thinking” answer, but he has championed “hills” as the correct answer, which swings things to the always@60mph interpretation.

[Gary_Robson]

[moderating]
Moved puzzle to the Game Room.
[/moderating]

[jharvey963]

Just out of curiosity, what would the difference in distance be between

1. the straight-line distance between A and B (as in, the distance a laser would travel between A and B, if the earth wasn’t in the way), and
2. the distance on the surface of the earth, assuming a completely “flat” (i.e., no hills or valleys) path between A and B.

I guess I’m asking the difference between the spherical path versus the straight-line path over a 60 mile run.

Thanks,
J.

[Hampshire]

If the distance of a mile is 5280 feet, and the extra three seconds gives us an extra 264 feet (=5544 feet total) you’d have to make a gross elevation change of 1690 feet over the couse of a mile. Is there anywhere that you can go a constant 60mph with that drastic of an elevation change?

[jharvey963]

@Hampshire

If the distance of a mile is 5280 feet, and the extra three seconds gives us an extra 264 feet (=5544 feet total) you’d have to make a gross elevation change of 1690 feet over the couse of a mile. Is there anywhere that you can go a constant 60mph with that drastic of an elevation change?

The 1690 feet would be over the course of 60 miles, not 1 mile, wouldn’t it?

J.

[Chessic_Sense]

@jharvey963

Just out of curiosity, what would the difference in distance be between

1. the straight-line distance between A and B (as in, the distance a laser would travel between A and B, if the earth wasn’t in the way), and
2. the distance on the surface of the earth, assuming a completely “flat” (i.e., no hills or valleys) path between A and B.

I guess I’m asking the difference between the spherical path versus the straight-line path over a 60 mile run.

Thanks,
J.

'bout 3 feet.

Assume the tunnel is 60 miles and the curve is longer and the Earth is a sphere.

Radius of Earth = 3963.17 mi
Half a tunnel = 30 mi.
Circumference of Earth = 2piR
Angle from beginning of tunnel to center = arcsin(30/3963.17)
Half of curve path/circumference = arcsin(30/3963.17)/2pi
Half of curve path = (arcsin(30/3963.17)/2pi)2piR miles
Curve path = (arcsin(30/3963.17)/2pi)2piR2 miles
Curve path = (arcsin(30/3963.17)/2pi)2piR25380 feet
Difference in paths = (arcsin(30/3963.17)/(2pi))2pi3963.1725280 - (60*5280)

= 3.0255 feet

[Chessic_Sense]

@Hampshire

If the distance of a mile is 5280 feet, and the extra three seconds gives us an extra 264 feet (=5544 feet total) you’d have to make a gross elevation change of 1690 feet over the couse of a mile. Is there anywhere that you can go a constant 60mph with that drastic of an elevation change?

Sure, if you only make up the distance in the last mile and only count going up a hill.

Imagine only one “triangle-ish” hill. The base would be 60 miles and the top would be 60+3/60 miles. What’s the center height?

((30+3/120)^2 - 30^2)^.5 = 1.22 miles = 6468 feet.

Excessive for just one hill, but not that far off. Easily accomplished if you use two hills.

[Hampshire]

@jharvey963

The 1690 feet would be over the course of 60 miles, not 1 mile, wouldn’t it?

J.

Ah crap, you’re right. 28 feet change per mile.

[Chessic_Sense]

@Hampshire

Ah crap, you’re right. 28 feet change per mile.

No, 216 feet per mile. But remember, that means you can go 108 up and 108 down and it’ll count just as well. That’s a 4% grade.

[Corresponder]

Thank you all for giving responses. I did mean that the car was travelling at a constant 60 mph from A to B without starting or stopping, and also was figuring on an average grade up or down of about 5%.