OT Ok @Lionjim (or even @IrishHerb) I need some math help

[s1uggo72]

as opposed to just general help.

Driving down the road I always wonder, suppose I drive at rate (mph) equal to the distance home, how long will it take to get home? so when I am 100 miles from home, I go 100 mph. When I am 86 miles from home, I am traveling at 86 mph. etc when I am 1 mile from home, I am traveling 1 mph.
I believe I'll get home, but I cant set this problem up. So can you help me get home? I think this is a Calculus 1 problem, but that was some time ago. TIA

 

[PSUFTG]

as opposed to just general help.

Driving down the road I always wonder, suppose I drive at rate (mph) equal to the distance home, how long will it take to get home? so when I am 100 miles from home, I go 100 mph. When I am 86 miles from home, I am traveling at 86 mph. etc when I am 1 mile from home, I am traveling 1 mph.
I believe I'll get home, but I cant set this problem up. So can you help me get home? I think this is a Calculus 1 problem, but that was some time ago. TIA

Calculus works - but you don't need it.
Just ask yourself, based on your premise, "How close do you need to be from Home, in order to get Home in 1 hour?" One mile away? One foot away? one nanometer away?

 

[s1uggo72]

Calculus works - but you don't need it.
Just ask yourself, based on your premise, "How close do you need to be from Home, in order to get Home in 1 hour?" One mile away? One foot away? one nanometer away?

I need to be home.

 

[LionJim]

There goes my afternoon. It’s likely a trick question, and I recall reading of a similar problem in a biography of Richard Feynman. Feynman, of course, solved it in a minute. (IMO, Feynman would have been one of the greatest mathematicians if he had gone that way. Heisenberg, too.)

 

[psuro]

as opposed to just general help.

Driving down the road I always wonder, suppose I drive at rate (mph) equal to the distance home, how long will it take to get home? so when I am 100 miles from home, I go 100 mph. When I am 86 miles from home, I am traveling at 86 mph. etc when I am 1 mile from home, I am traveling 1 mph.
I believe I'll get home, but I cant set this problem up. So can you help me get home? I think this is a Calculus 1 problem, but that was some time ago. TIA

There is an equation in there somewhere....just have to find it.

There goes my afternoon.

 

[PSUFTG]

I need to be home.

Based on your premise ("I drive at rate (mph) equal to the distance home"), will you ever be close enough to Home to get there in 1 hour? Therein lies your answer. And it is quite simple, really, if you just "figure it" it doesn't even require any math.

 

[s1uggo72]

There goes my afternoon. It’s likely a trick question, and I recall reading of a similar problem in a biography of Richard Feynman. Feynman, of course, solved it in a minute. (IMO, Feynman would have been one of the greatest mathematicians if he had gone that way. Heisenberg, too.)

FTR I dont intend it to be a trick question, but a practical one. I dont think the fact that I am constantly decelerating has anything to do with it. I can figure out how to express that D= V, that is 100 miles away means I am going 100 mph, while 23 miles away I am going 23 mph

I think of this problem every time I am on the 'road' going somewhere.

 

[s1uggo72]

Based on your premise, will you ever be close enough to Home to get there in 1 hour? Therein lies your answer.

huh? my rate will always be equal to my distance from home, but we should be able to divide up a finite distance into an infinite amount of intervals.

 

[BobPSU92]

I never did the MATHS. . Instead, I asked my dad “Are we there yet?” until he smacked the sh|t out of me.

 

[91Joe95]

as opposed to just general help.

Driving down the road I always wonder, suppose I drive at rate (mph) equal to the distance home, how long will it take to get home? so when I am 100 miles from home, I go 100 mph. When I am 86 miles from home, I am traveling at 86 mph. etc when I am 1 mile from home, I am traveling 1 mph.
I believe I'll get home, but I cant set this problem up. So can you help me get home? I think this is a Calculus 1 problem, but that was some time ago. TIA

Deceleration as a function of distance?

deceleration=(sf^2-si^2)/2d

sf = final speed
si = initial speed
d = distance

How to Calculate Deceleration

Deceleration means slowing down, the opposite of acceleration. Deceleration may be calculated using either the time or the distance over which the change in speed occurs. Deceleration may be expressed in gravitational units (G).

sciencing.com

Then solve for t in this equation?

Acceleration to Distance Calculator - Calculator Academy

Enter the initial velocity, acceleration, and total time into the calculator to convert the acceleration into total distance.

calculator.academy

 

[91Joe95]

I never did the MATHS. . Instead, I asked my dad “Are we there yet?” until he smacked the sh|t out of me.

Was it his favorite travel game?

 

[s1uggo72]

Deceleration as a function of distance?

deceleration=(sf^2-si^2)/2d

sf = final speed
si = initial speed
d = distance

How to Calculate Deceleration

Deceleration means slowing down, the opposite of acceleration. Deceleration may be calculated using either the time or the distance over which the change in speed occurs. Deceleration may be expressed in gravitational units (G).

sciencing.com

Then solve for t in this equation?

Acceleration to Distance Calculator - Calculator Academy

Enter the initial velocity, acceleration, and total time into the calculator to convert the acceleration into total distance.

calculator.academy

I dont think cleration (either DE or Ac) has anything to do with it. your velocity (in MPH) =D.

 

[CamelbackPSU]

Definitely a calculus problem and that part of me is long gone. But using Excel, and assuming you could go 100 mph for the entire time it took to go 1 mile, then 99 mph for the time it took to go 1 mile & summing the 100 times (ie time in minutes= 60*1 mile/mph) you get 311.2427 minutes, with the last mile taking 60 minutes. If you took the last mile and started with 1 mph but divided it up by .01 mile segments that last mile would take the same 311.2427 minutes. So maybe you would never get home. But I have trouble going less than 15 mph so I would beat you to the destination.

 

[Waaaaaaaany]

We need more info, like at what point are you going so infinitely slow that you decide to just get out and walk home

 

[IrishHerb]

You'd never get home. When you're over 80 miles from home, you'd be stopped by the Penn State Highway Patrol.

 

[s1uggo72]

You'd never get home. When you're over 80 miles from home, you'd be stopped by the Penn State Highway Patrol.

good thing I was in Kentucky, when I was doing 80, I kept getting passed!

 

[LionJim]

You'll never get home. Suppose x(t) is the distance from home. Then y(t)=100-x(t) is the total distance travelled. From this, the velocity is y'=-x'. But we've already said that the velocity y' is x, so we have x=-x', equivalent to x'=-x, which gives x=ce^(-t). Since x(0)=100, c=100 and so x(t)=100e^(-t). Since the RHS of this equation is never zero, x(t) is never zero, so you'll never get home.

 

[s1uggo72]

You'll never get home. Suppose x(t) is the distance from home. Then y(t)=100-x(t) is the total distance travelled. From this, the velocity is y'=-x'. But we've already said that the velocity y' is x, so we have x=-x', equivalent to x'=-x, which gives x=ce^(-t). Since x(0)=100, c=100 and so x(t)=100e^(-t). Since the RHS of this equation is never zero, x(t) is never zero, so you'll never get home.

Thanks. And it makes me feel better it has taken you so long!!!
I just figured that since it was a finite distance it could be divided into into infinite intervals, like the old walking to the wall paradox. To get to the wall , walk half the distance, now walk 1/2 of that distance, etc so there’s always 1/2 of whatever to go, yet I know I can get to the wall

 

[Nohow]

You'll never get home. Suppose x(t) is the distance from home. Then y(t)=100-x(t) is the total distance travelled. From this, the velocity is y'=-x'. But we've already said that the velocity y' is x, so we have x=-x', equivalent to x'=-x, which gives x=ce^(-t). Since x(0)=100, c=100 and so x(t)=100e^(-t). Since the RHS of this equation is never zero, x(t) is never zero, so you'll never get home.

So a variation on Zeno’s paradox?

 

[s1uggo72]

So a variation on Zeno’s paradox?

Isn’t that is what calculus is based? Dividing a finite space into infinite amount of sums

 

[PSUJam]

We need more info, like at what point are you going so infinitely slow that you decide to just get out and walk home

When I told my mother I was bored as a kid, she would tell me to go play in traffic. True story.

 

[fairgambit]

 

[LionJim]

So a variation on Zeno’s paradox?

What fools people with Zeno's Paradox is the fact that they are dealing with a convergent infinite sum, 1/2 + 1/4 + 1/8 + 1/16 + ... = 1. Think about it: if you walk 10 feet per second, how long would it take you to walk 10 feet? One second, which is 1 = 1/2 + 1/4 + 1/8 + ...

With Sluggo's problem it takes a tad less than 41.6 minutes ( log(2) hours) to drive the first half, from the 100 mile mark to the 50 mile mark, the same 41.6 minutes to drive from the 50 mile mark to the 25 mile mark, the same 41.6 minutes to drive from the 25 mile mark to the 12.5 = 25/2 mile mark, and so on. Add up these times and it's a divergent infinite sum.

 

[s1uggo72]

What fools people with Zeno's Paradox is the fact that they are dealing with a convergent infinite sum, 1/2 + 1/4 + 1/8 + 1/16 + ... = 1. Think about it: if you walk 10 feet per second, how long would it take you to walk 10 feet? One second, which is 1 = 1/2 + 1/4 + 1/8 + ...

With Sluggo's problem it takes a tad less than 41.6 minutes ( log(2) hours) to drive the first half, from the 100 mile mark to the 50 mile mark, the same 41.6 minutes to drive from the 50 mile mark to the 25 mile mark, the same 41.6 minutes to drive from the 25 mile mark to the 12.5 = 25/2 mile mark, and so on. Add up these times and it's a divergent infinite sum.

Can we call this Sluggos paradox??

 

[TiogaLion]

It's a series with: Sum of (1/n x 3600 sec); from n=1, 100; where n = both the velocity and distance from home with the summation being the number of seconds it will take to get home. This of course assumes deceleration is a step function of 1 at each mile marker.

 

[step.eng69]

I never did the MATHS. . Instead, I asked my dad “Are we there yet?” until he smacked the sh|t out of me.

Yep,
I can see a quick slap to the back of the head could relieve some built up tension you had.

 

[fairgambit]

 

[PSU12046]

View attachment 198052

Anytime you can include Danica McKellar in a thread, it's a bonus! Fair FTW!

 

[LionJim]

You'll never get home. Suppose x(t) is the distance from home. Then y(t)=100-x(t) is the total distance travelled. From this, the velocity is y'=-x'. But we've already said that the velocity y' is x, so we have x=-x', equivalent to x'=-x, which gives x=ce^(-t). Since x(0)=100, c=100 and so x(t)=100e^(-t). Since the RHS of this equation is never zero, x(t) is never zero, so you'll never get home.

Here's an explanation which doesn't use calculus. Remember that Sluggo is driving from Mile Marker 100 To MM 0 and his speed at MM X is exactly Xmph. What I'm going to do is set up a sequence of examples of a car arriving home before Sluggo, which we know will happen because at any specific Mile Marker Sluggo is never driving faster than the example car.

Example 1. The car is driven 100mph the whole time. Total travel time: 1 hour. It would thus take Sluggo more than one hour to get home.
Example 2. The car is driven 100mph for the first 50 miles and then 50mph for the remaining 50 miles. (Ignore the discontinuity.) Total travel time: 1.5 hours.
Example 3: First 50 miles at 100mph. The next 25 miles at 50mph. The final 25 miles at 25 mph. Total travel time 2 hours.
Example 4. First 50 miles at 100mph. The next 25 miles at 50mph. The next 12.5 miles at 25mph. The final 12.5 miles at 12.5mph. Total travel time 2.5 hours.

Example N: Total travel time (N+1)/2 hours.

You keep doing this and in each new example the travel time is a half-hour longer than that of the preceding example, and in each example the car will arrive home before Sluggo. The travel time of the examples increases without bound and in each example the car beats Sluggo home. So, to illustrate, I can find an example which takes 123473834627857356 hours to drive the 100 miles and Sluggo is still in the rear-view. Pick any positive finite number K you want and you'll find an example that takes more than K hours to get home and Sluggo is sluggoing behind. If Sluggo made it home in M hours, then you'll be able to find an example which took more than M+1 hours and sluggo is still not home, a contradiction. Thus Sluggo never makes it home.

 

[PSUAXE70]

Here's an explanation which doesn't use calculus. Remember that Sluggo is driving from Mile Marker 100 To MM 0 and his speed at MM X is exactly Xmph. What I'm going to do is set up a sequence of examples of a car arriving home before Sluggo, which we know will happen because at any specific Mile Marker Sluggo is never driving faster than the example car.

Example 1. The car is driven 100mph the whole time. Total travel time: 1 hour. It would thus take Sluggo more than one hour to get home.
Example 2. The car is driven 100mph for the first 50 miles and then 50mph for the remaining 50 miles. (Ignore the discontinuity.) Total travel time: 1.5 hours.
Example 3: First 50 miles at 100mph. The next 25 miles at 50mph. The final 25 miles at 25 mph. Total travel time 2 hours.
Example 4. First 50 miles at 100mph. The next 25 miles at 50mph. The next 12.5 miles at 25mph. The final 12.5 miles at 12.5mph. Total travel time 2.5 hours.

Example N: Total travel time (N+1)/2 hours.

You keep doing this and in each new example the travel time is a half-hour longer than that of the preceding example, and in each example the car will arrive home before Sluggo. The travel time of the examples increases without bound and in each example the car beats Sluggo home. So, to illustrate, I can find an example which takes 123473834627857356 hours to drive the 100 miles and Sluggo is still in the rear-view. Pick any positive finite number K you want and you'll find an example that takes more than K hours to get home and Sluggo is sluggoing behind. If Sluggo made it home in M hours, then you'll be able to find an example which took more than M+1 hours and sluggo is still not home, a contradiction. Thus Sluggo never makes it home.

The time to travel one mile at 100 mph is finite. So is the the time to travel one mile at 99 mph. Add the 100 finite numbers and you will get a finite sum. I’m guessing Camelback is right.

 

[PSUFTG]

The time to travel one mile at 100 mph is finite. So is the the time to travel one mile at 99 mph. Add the 100 finite numbers and you will get a finite sum. I’m guessing Camelback is right.

Yep. It's a specific question that takes about 10 seconds of thinking to answer.

 

[LionJim]

Based on your premise ("I drive at rate (mph) equal to the distance home"), will you ever be close enough to Home to get there in 1 hour? Therein lies your answer. And it is quite simple, really, if you just "figure it" it doesn't even require any math.

I see your point now. Very nice. No matter where he is, Sluggo will always be more than an hour away from home. If he is x miles away and he maintains his speed (x mph) at that moment, it will take him exactly an hour to get home. Since he doesn’t maintain his speed, he’ll always be more than an hour away. Ten miles out, he’s more than an hour away from home. Ten inches out, he’s more than an hour away from home. He’ll never get home. Very nice.

 

[IrishHerb]

View attachment 198052

If you want a really hot math teacher, here's one ... Danica is nice, but this one is ...

 

[TiogaLion]

I see your point now. Very nice. No matter where he is, Sluggo will always be more than an hour away from home. If he is x miles away and he maintains his speed (x mph) at that moment, it will take him exactly an hour to get home. Since he doesn’t maintain his speed, he’ll always be more than an hour away. Ten miles out, he’s more than an hour away from home. Ten inches out, he’s more than an hour away from home. Very nice.

The last mile will take him one hour to get home. However, the OP asked how long to get home from x miles going x mph if he decreases his speed by one mph after each mile. The length of the trip home is the summation of the time it takes for each of the x miles in the trip.

 

[Woodpecker]

Thomas Wolfe would argue that you can't go home again.

 

[s1uggo72]

The last mile will take him one hour to get home. However, the OP asked how long to get home from x miles going x mph if he decreases his speed by one mph after each mile. The length of the trip home is the summation of the time it takes for each of the x miles in the trip.

not really, because when I am 1/2 mile from home, I'll only be going 1/2 mph, still have an hour to go!! 1/4 mile the same etc.

 

[TiogaLion]

not really, because when I am 1/2 mile from home, I'll only be going 1/2 mph, still have an hour to go!! 1/4 mile the same etc.

The problem as stated by the OP makes me believe he intended a step function at each mile marker. The problem you stated means he'll never get home.

 

[LionJim]

The last mile will take him one hour to get home. However, the OP asked how long to get home from x miles going x mph if he decreases his speed by one mph after each mile. The length of the trip home is the summation of the time it takes for each of the x miles in the trip.

No, his speed decreases continuously. He’s not driving the first mile at 100 mph, the second mile at 99 mph, and so on. For example, at mile marker 92.5 he’s going 92.5 mph, not 93 mph, as you say. If it’s a step function, you’re correct. If one assumes that it’s a continuously decreasing function, he’ll never get home.

 

[Shadow99]

There goes my afternoon. It’s likely a trick question, and I recall reading of a similar problem in a biography of Richard Feynman. Feynman, of course, solved it in a minute. (IMO, Feynman would have been one of the greatest mathematicians if he had gone that way. Heisenberg, too.)

idk, I would think that whether or not Heisenberg would have been a great mathematician is uncertain.

 

[s1uggo72]

The problem as stated by the OP makes me believe he intended a step function at each mile marker. The problem you stated means he'll never get home.

since I am the OP, I intended it to be whatever MM I am at, that would be my rate. so 66.5 miles from home, I'll be going 66.5 mph. etc.
see I think I get home, as a finite space can be divided into an infinite amount of intervals, at least thats what I remember from calculus.