ATOT Maths Challenge

[DisgruntledVirus]

Originally posted by: cubby1223
x=distance when the first walker was picked up
y=distance when the biker began walking

65-x=y
x @ 5kmph + y @ 50kmph = (2y+(65-2x)) @ 50kmph

whoever wants to work it out, I don't care. Only thing I'd like to know is, the motorbike always runs at a constant 50kmph, must have been rough when the one was shoved off the bike at such high velocity. 😉



edit - I'm failing to see how this is considered "higher" math. :confused;

Well you know those politicians these days, they don't want no child left behind even though they were left behind in 4th grade.

 

[Quasmo]

They arrive on Thursday.

 

[Kev]

Maths? Full marks? Kilometers? Go away I hate you.

 

[OdiN]

Originally posted by: cubby1223
x=distance when the first walker was picked up
y=distance when the biker began walking

65-x=y
x @ 5kmph + y @ 50kmph = (2y+(65-2x)) @ 50kmph

whoever wants to work it out, I don't care. Only thing I'd like to know is, the motorbike always runs at a constant 50kmph, must have been rough when the one was shoved off the bike at such high velocity. 😉



edit - I'm failing to see how this is considered "higher" math. 😕

I think that the instantaneous acceleration from 0 to 50KMPH is going to kind of hurt.

 

[sdifox]

Originally posted by: OdiN

Originally posted by: cubby1223
x=distance when the first walker was picked up
y=distance when the biker began walking

65-x=y
x @ 5kmph + y @ 50kmph = (2y+(65-2x)) @ 50kmph

whoever wants to work it out, I don't care. Only thing I'd like to know is, the motorbike always runs at a constant 50kmph, must have been rough when the one was shoved off the bike at such high velocity. 😉



edit - I'm failing to see how this is considered "higher" math. 😕

I think that the instantaneous acceleration from 0 to 50KMPH is going to kind of hurt.

Inertial dampening 🙂, the wonderful Star Trek physics.

 

[OdiN]

Originally posted by: sdifox

Originally posted by: OdiN

Originally posted by: cubby1223
x=distance when the first walker was picked up
y=distance when the biker began walking

65-x=y
x @ 5kmph + y @ 50kmph = (2y+(65-2x)) @ 50kmph

whoever wants to work it out, I don't care. Only thing I'd like to know is, the motorbike always runs at a constant 50kmph, must have been rough when the one was shoved off the bike at such high velocity. 😉



edit - I'm failing to see how this is considered "higher" math. 😕

I think that the instantaneous acceleration from 0 to 50KMPH is going to kind of hurt.

Inertial dampening 🙂, the wonderful Star Trek physics.

Yeah and now that I think about it, it would hurt more when you are going 50KMPH in one direction, then instantly are going 50KMPH in the opposite direction.

 

[JujuFish]

Originally posted by: Barack Obama

Originally posted by: KnickNut3
11:06am

Explain please. I'm not going to go check my answers without good reason.

Didn't think very deeply into it, but seems accurate to me. The two walkers both have to walk the same distance, so it's pretty simple to set up an equation.

X = the distance either walker has to walk

Time it takes for the bike rider:
(65 - x)/50 + (65 - 2·x)/50 + (65 - x)/50

Time it takes for either walker:
x/5 + (65 - x)/50

Equating those two gives x=10. Plugging that back into one side of the equation gives 3.1 hours, or 11:06.

 

[alkemyst]

can they CL the motorbike in trade for a faster ride?

 

[cubby1223]

Originally posted by: DisgruntledVirus

Originally posted by: cubby1223
edit - I'm failing to see how this is considered "higher" math. :confused;

Well you know those politicians these days, they don't want no child left behind even though they were left behind in 4th grade.

Perhaps this is "higher" math for those who began in "new" math. Geez, my cousin's kids are in a district that teaches that stuff, I was looking over the homework once... boggles my mind how the kids learn *anything*. 🙁 They have no clue what the actual arithmetic is, all they know is how to rearrange objects as the instructions tell them to.

It's preparing them for jobs where all they have to do is punch buttons on the cash register at McDonald's, and trust the readout is correct. 😉

 

[cubby1223]

Originally posted by: OdiN
Yeah and now that I think about it, it would hurt more when you are going 50KMPH in one direction, then instantly are going 50KMPH in the opposite direction.

Ever play the Half-Life add-on Portal? 😀 Speedy thing go in, speedy thing come out.

 

[JujuFish]

Originally posted by: cubby1223

Originally posted by: OdiN
Yeah and now that I think about it, it would hurt more when you are going 50KMPH in one direction, then instantly are going 50KMPH in the opposite direction.

Ever play the Half-Life add-on Portal? 😀 Speedy thing go in, speedy thing come out.

Portal is not a Half-Life add-on.

Edit: I may have misread that. Is there an actual add-on for Half-Life that adds the Portal gun?

 

[OdiN]

Originally posted by: cubby1223

Originally posted by: OdiN
Yeah and now that I think about it, it would hurt more when you are going 50KMPH in one direction, then instantly are going 50KMPH in the opposite direction.

Ever play the Half-Life add-on Portal? 😀 Speedy thing go in, speedy thing come out.

Technically you're still travelling in the same "direction", you just are going through a dimensional rift of some sort.

Whether or not you could survive that is a debate in and of itself.

 

[FleshLight]

65/50 + 2x/50 = x/50 + (65-x)/5

65/50 + x/50 = (65-x)/5

x = 53.1818

3 hours and 25.8 minutes = 11:25:8/60

 

[ivan2]

im writing down the equations before i throw them away:

X = distance girl 1 walked
Y = distance girl 2 walked (the one who got projected off the bike at 50km/h and immediately slowed down to 5km/h using star trek technology)

X/5 + (65-X)/50=Y/5

((65-Y) + (65-X-Y)) / 50 = X/5

solve for X, Y then use the distance the bike traveled to calculate the time.

comes out to roughly 2.945 hours.

 

[Kyteland]

There are three times that are important, call them t1, t2 and t3. t1 is when Mary got off the bike. At t1 the distances traveled by each person are:
L(t1) = 50*t1
M(t1) = 50*t1
N(t1) = 5*t1

The next important time is t2, when Nora gets on the bike. At t2 the distances traveled by each person are:
L(t2) = L(t1) - 50*(t2-t1) = 100*t1 - 50*t2
M(t2) = M(t1) + 5*(t2-t1) = 45*t1 + 5*t2
N(t2) = N(t1) + 5*(t2-t1) = 5*t2

The next important time is t3, when they all reach their destinations. At t3 the distances traveled by each person are:
L(t3) = L(t2) + 50*(t3-t2) = 100*t1 - 100*t2 + 50*t3
M(t3) = M(t2) + 5*(t3-t2) = 45*t1 + 5*t3
N(t3) = N(t2) + 50*(t3-t2) = 50*t3 - 45*t2

We also know that at t3 they have all traveled 65km which gives a system of equations
65 = 100*t1 - 100*t2 + 50*t3
65 = 45*t1 + 5*t3
65 = 50*t3 - 45*t2

Solving the system gives
t1 = 1.1h = 66m
t2 = 2h = 120m
t3 = 3.1h = 186m

So the final answer is 11:06AM

 

[chuckywang]

I get 11:06am.

 

[chuckywang]

Originally posted by: ivan2
im writing down the equations before i throw them away:

X = distance girl 1 walked
Y = distance girl 2 walked (the one who got projected off the bike at 50km/h and immediately slowed down to 5km/h using star trek technology)

X/5 + (65-X)/50=Y/5

((65-Y) + (65-X-Y)) / 50 = X/5

solve for X, Y then use the distance the bike traveled to calculate the time.

comes out to roughly 2.945 hours.

Your equations are borked. You forgot to take into account the bike had to backtrack, pick someone up, and then recover that distance.

 

[ivan2]

Originally posted by: chuckywang

Originally posted by: ivan2
im writing down the equations before i throw them away:

X = distance girl 1 walked
Y = distance girl 2 walked (the one who got projected off the bike at 50km/h and immediately slowed down to 5km/h using star trek technology)

X/5 + (65-X)/50=Y/5

((65-Y) + (65-X-Y)) / 50 = X/5

solve for X, Y then use the distance the bike traveled to calculate the time.

comes out to roughly 2.945 hours.

Your equations are borked. You forgot to take into account the bike had to backtrack, pick someone up, and then recover that distance.

the second equation covered that. i just need to go back to the TI89 to see if it's the equation or is me that is wrong LOL. I think the 11:06 is a reasonable time.

 

[chuckywang]

Originally posted by: ivan2

Originally posted by: chuckywang

Originally posted by: ivan2
im writing down the equations before i throw them away:

X = distance girl 1 walked
Y = distance girl 2 walked (the one who got projected off the bike at 50km/h and immediately slowed down to 5km/h using star trek technology)

X/5 + (65-X)/50=Y/5

((65-Y) + (65-X-Y)) / 50 = X/5

solve for X, Y then use the distance the bike traveled to calculate the time.

comes out to roughly 2.945 hours.

Your equations are borked. You forgot to take into account the bike had to backtrack, pick someone up, and then recover that distance.

the second equation covered that. i just need to go back to the TI89 to see if it's the equation or is me that is wrong LOL. I think the 11:06 is a reasonable time.

I was talking about the first equation, I didn't look at your second one, but now that I did, your second equation is correct. The problem with the first equation is that you're not equating the same things:

Left hand side = time girl 1 walked, then time girl 1 was on bike to destination. therefore, this is the total time the girls spent on the bike
Right hand side = time girl 2 walked, which is not the same as the total time she spent getting to the destination

 

[ivan2]

Originally posted by: chuckywang

Originally posted by: ivan2

Originally posted by: chuckywang

Originally posted by: ivan2
im writing down the equations before i throw them away:

X = distance girl 1 walked
Y = distance girl 2 walked (the one who got projected off the bike at 50km/h and immediately slowed down to 5km/h using star trek technology)

X/5 + (65-X)/50=Y/5

((65-Y) + (65-X-Y)) / 50 = X/5

solve for X, Y then use the distance the bike traveled to calculate the time.

comes out to roughly 2.945 hours.

Your equations are borked. You forgot to take into account the bike had to backtrack, pick someone up, and then recover that distance.

the second equation covered that. i just need to go back to the TI89 to see if it's the equation or is me that is wrong LOL. I think the 11:06 is a reasonable time.

I was talking about the first equation, I didn't look at your second one, but now that I did, your second equation is correct.

ahhh ur right.

 

[JujuFish]

Originally posted by: ivan2

Originally posted by: chuckywang

Originally posted by: ivan2
im writing down the equations before i throw them away:

X = distance girl 1 walked
Y = distance girl 2 walked (the one who got projected off the bike at 50km/h and immediately slowed down to 5km/h using star trek technology)

X/5 + (65-X)/50=Y/5

((65-Y) + (65-X-Y)) / 50 = X/5

solve for X, Y then use the distance the bike traveled to calculate the time.

comes out to roughly 2.945 hours.

Your equations are borked. You forgot to take into account the bike had to backtrack, pick someone up, and then recover that distance.

the second equation covered that. i just need to go back to the TI89 to see if it's the equation or is me that is wrong LOL. I think the 11:06 is a reasonable time.

Considering both girls walked the same distance, your first equation is definitely wrong.

 

[ivan2]

Originally posted by: JujuFish

Considering both girls walked the same distance, your first equation is definitely wrong.

that's the part of your solution that I don't get if you can explain that would be nice.

 

[SSSnail]

Don't know, because there was a mall on the way.

 

[JujuFish]

Originally posted by: ivan2

Originally posted by: JujuFish

Considering both girls walked the same distance, your first equation is definitely wrong.

that's the part of your solution that I don't get if you can explain that would be nice.

I'm not sure what there is to "get". Make a diagram if you need to. Since they both walk at the same rate and everyone starts and ends at the same time, they both end up walking 2 hours for 10 kilometers.

 

[Fenixgoon]

Originally posted by: Kyteland
There are three times that are important, call them t1, t2 and t3. t1 is when Mary got off the bike. At t1 the distances traveled by each person are:
L(t1) = 50*t1
M(t1) = 50*t1
N(t1) = 5*t1

The next important time is t2, when Nora gets on the bike. At t2 the distances traveled by each person are:
L(t2) = L(t1) - 50*(t2-t1) = 100*t1 - 50*t2
M(t2) = M(t1) + 5*(t2-t1) = 45*t1 + 5*t2
N(t2) = N(t1) + 5*(t2-t1) = 5*t2

The next important time is t3, when they all reach their destinations. At t3 the distances traveled by each person are:
L(t3) = L(t2) + 50*(t3-t2) = 100*t1 - 100*t2 + 50*t3
M(t3) = M(t2) + 5*(t3-t2) = 45*t1 + 5*t3
N(t3) = N(t2) + 50*(t3-t2) = 50*t3 - 45*t2

We also know that at t3 they have all traveled 65km which gives a system of equations
65 = 100*t1 - 100*t2 + 50*t3
65 = 45*t1 + 5*t3
65 = 50*t3 - 45*t2

Solving the system gives
t1 = 1.1h = 66m
t2 = 2h = 120m
t3 = 3.1h = 186m

So the final answer is 11:06AM

hooray matrix math. i was too lazy to solve it, i knew how to set up the problem though 🙂