Mixing
Distances and speeds
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Alice and Bob started to walk towards each other’s home and then back to theirs, with steady speeds. Alice passed by a bus station at 25 m away from her home, while at the same time Bob was passing by an abandoned old car. Afterwards, they met at 55 m away from Bob’s home and then they met again at 85 m from Alice’s home.
What is the distance between the bus station and the abandoned car?
This must be an easy problem but it's been years since I was dealing with maths & physics!
Trying to set up the equations:
If x is the required distance, S the distance between Alice's & Bob's homes and $t_1$ time units passed since they started walking and until they first encountered the bus and the car, also $V_1$ and $V_2$ Alice's and Bob's speed respectively:
$V_1.t_1 + x + V_2.t_1 = S$
$V_1.t_1 = 25$
Then they met after $t_2$ time units:
$V_1.t_2 + V_2.t_2 = S$ and
$V_2.t_2 = 55$
Finally they met again after $t_3$ time units:
$V_1.t_3 + V_2.t_3 = 3S$ (because each one traveled S plus his part of S).
Also
$V_2.t_3 = S + 85$
Then what? Apparently I am missing one equation!
Thank you very much!
arithmetic
asked Jul 22 at 11:33
Nhung Huyen
12410
add a comment |Â
up vote
11
down vote
favorite
Alice and Bob started to walk towards each other’s home and then back to theirs, with steady speeds. Alice passed by a bus station at 25 m away from her home, while at the same time Bob was passing by an abandoned old car. Afterwards, they met at 55 m away from Bob’s home and then they met again at 85 m from Alice’s home.
What is the distance between the bus station and the abandoned car?
This must be an easy problem but it's been years since I was dealing with maths & physics!
Trying to set up the equations:
If x is the required distance, S the distance between Alice's & Bob's homes and $t_1$ time units passed since they started walking and until they first encountered the bus and the car, also $V_1$ and $V_2$ Alice's and Bob's speed respectively:
$V_1.t_1 + x + V_2.t_1 = S$
$V_1.t_1 = 25$
Then they met after $t_2$ time units:
$V_1.t_2 + V_2.t_2 = S$ and
$V_2.t_2 = 55$
Finally they met again after $t_3$ time units:
$V_1.t_3 + V_2.t_3 = 3S$ (because each one traveled S plus his part of S).
Also
$V_2.t_3 = S + 85$
Then what? Apparently I am missing one equation!
Thank you very much!
arithmetic
asked Jul 22 at 11:33
Nhung Huyen
12410
I'm wondering if it's possible to solve this problem in a geometrical way, e.g. with a time distance diagram.
– Eric Duminil
Jul 22 at 20:25
add a comment |Â
up vote
11
down vote
favorite
up vote
11
down vote
favorite
Alice and Bob started to walk towards each other’s home and then back to theirs, with steady speeds. Alice passed by a bus station at 25 m away from her home, while at the same time Bob was passing by an abandoned old car. Afterwards, they met at 55 m away from Bob’s home and then they met again at 85 m from Alice’s home.
What is the distance between the bus station and the abandoned car?
This must be an easy problem but it's been years since I was dealing with maths & physics!
Trying to set up the equations:
If x is the required distance, S the distance between Alice's & Bob's homes and $t_1$ time units passed since they started walking and until they first encountered the bus and the car, also $V_1$ and $V_2$ Alice's and Bob's speed respectively:
$V_1.t_1 + x + V_2.t_1 = S$
$V_1.t_1 = 25$
Then they met after $t_2$ time units:
$V_1.t_2 + V_2.t_2 = S$ and
$V_2.t_2 = 55$
Finally they met again after $t_3$ time units:
$V_1.t_3 + V_2.t_3 = 3S$ (because each one traveled S plus his part of S).
Also
$V_2.t_3 = S + 85$
Then what? Apparently I am missing one equation!
Thank you very much!
arithmetic
asked Jul 22 at 11:33
Nhung Huyen
12410
Alice and Bob started to walk towards each other’s home and then back to theirs, with steady speeds. Alice passed by a bus station at 25 m away from her home, while at the same time Bob was passing by an abandoned old car. Afterwards, they met at 55 m away from Bob’s home and then they met again at 85 m from Alice’s home.
What is the distance between the bus station and the abandoned car?
This must be an easy problem but it's been years since I was dealing with maths & physics!
Trying to set up the equations:
If x is the required distance, S the distance between Alice's & Bob's homes and $t_1$ time units passed since they started walking and until they first encountered the bus and the car, also $V_1$ and $V_2$ Alice's and Bob's speed respectively:
$V_1.t_1 + x + V_2.t_1 = S$
$V_1.t_1 = 25$
Then they met after $t_2$ time units:
$V_1.t_2 + V_2.t_2 = S$ and
$V_2.t_2 = 55$
Finally they met again after $t_3$ time units:
$V_1.t_3 + V_2.t_3 = 3S$ (because each one traveled S plus his part of S).
Also
$V_2.t_3 = S + 85$
Then what? Apparently I am missing one equation!
Thank you very much!
arithmetic
asked Jul 22 at 11:33
Nhung Huyen
12410
asked Jul 22 at 11:33
Nhung Huyen
12410
asked Jul 22 at 11:33
Nhung Huyen
12410
asked Jul 22 at 11:33
Nhung Huyen
12410
12410
I'm wondering if it's possible to solve this problem in a geometrical way, e.g. with a time distance diagram.
– Eric Duminil
Jul 22 at 20:25
add a comment |Â
I'm wondering if it's possible to solve this problem in a geometrical way, e.g. with a time distance diagram.
– Eric Duminil
Jul 22 at 20:25
I'm wondering if it's possible to solve this problem in a geometrical way, e.g. with a time distance diagram.
– Eric Duminil
Jul 22 at 20:25
I'm wondering if it's possible to solve this problem in a geometrical way, e.g. with a time distance diagram.
– Eric Duminil
Jul 22 at 20:25
add a comment |Â
1 Answer
1
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oldest
votes
up vote
10
down vote
accepted
You have $V_1t_2=S-55 $, $V_2t_2=55$, $V_1t_3=2S-85$, $V_2t_3=S+85$.
Hence $$fracV_1V_2=fracS-5555=frac2S-85S+85 $$ Thus
$$55(2S-85)=(S+85)(S-55)=S^2+30S-85cdot 55 $$ $$Rightarrow
110S=S^2+30SRightarrow S=80 $$ (assuming $S>0$). In particular
$$fracV_1V_2=frac2555 $$ and so the distance between the bus
and the car is $$80-25-25cdot frac5525=0 $$
EDIT: This solution is not acceptable (and not the intended solution) because if the distance between Alice's and Bob's house is $80$, and they are only moving between each other's houses, they cannot meet $85$ meters from Alice's house. The equation
$$fracV_1V_2=fracS-5555 $$
is correct because the first meeting will necessarily occur before both Alice and Bob have completed their first journey. However, for the second meeting there are other possible cases:
1) It could be that they meet again before Alice has reached Bob's house in her first trip. In this case, the second meeting occurs when in total Alice has travelled $85$ meters and Bob has travelled $S+85$ meters. We obtain
$$ fracV_1V_2=fracS-5555=frac85S+85Rightarrow S^2+30S-2cdot 55cdot 85=0$$
a quadratic equation whose positive root is
$$ S=5(sqrt383-3)approx 82.85$$
Since $82.85<85$, this solution is also unacceptable!
2) The remaining case is the one where they meet again before Bob has reached Alice's house in his first trip. Here at the second meeting in total Bob has travelled $S-85$ meters, while Alice has travelled $2S-85$ meters. We get
$$fracV_1V_2=fracS-5555=frac2S-85S-85 Rightarrow S^2-250S+2cdot 55cdot 85=0$$
a quadratic equation with roots $5(25pm sqrt251)$, of which only
$$S=5(25+sqrt251)approx 204 > 85 $$
is acceptable because $5(25-sqrt251)<85$. The distance between the bus and the car is then
beginalign*x&=S-25-25cdotfracV_2V_1=S-25-25frac55S-55=frac(S-25)(S-55)-25cdot 55S-55=fracS^2-80SS-55=\
&=frac250S-2cdot 55 cdot 85-80SS-55=frac170(S-55)S-55=170
endalign*
This is the intended solution of the problem. Indeed, there are no other possible cases: at the second meeting either Alice or Bob must have completed at least one journey, but none of them can have completed more than one, else there would have been another meeting inbetween.
answered Jul 22 at 13:08
Lorenzo Quarisa
2,284314
Lorenzo Quarisa: in the last row of your solution, how do you substitute $V_2.t_1$ with $$25.frac 5525$$?
– Nhung Huyen
Jul 22 at 19:51
1
OK so the result does not make any sense because if the total distance between the two houses is 80 m, this means that when Alice passed by the bus station (at 25 m from her house), at that moment, Bob was passing by the old car which was (80-25) = 55 m from his house. Therefore until their first meeting (at 55 m away from Bob's house), Bob did not move at all.
– Pradeep Suny
Jul 23 at 7:55
1
I agree that the solution is incorrect, not much because their first meeting is supposed to be at the bus+abandoned car, but rather because if they are only moving between each other's homes and the distance between them is $S=80$, they cannot meet at a distance of $85$ from Alice's house. Therefore the situation OP considered does not lead to any solutions, but there are other cases. I am looking into them now.
– Lorenzo Quarisa
Jul 23 at 8:22
1
@PradeepSuny I have updated my answer with the correct solution (I believe).
– Lorenzo Quarisa
Jul 23 at 9:09
1
Yes this one is correct and the result is 170. I hope the OP also finds it useful.
– Pradeep Suny
Jul 23 at 11:16
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
You have $V_1t_2=S-55 $, $V_2t_2=55$, $V_1t_3=2S-85$, $V_2t_3=S+85$.
Hence $$fracV_1V_2=fracS-5555=frac2S-85S+85 $$ Thus
$$55(2S-85)=(S+85)(S-55)=S^2+30S-85cdot 55 $$ $$Rightarrow
110S=S^2+30SRightarrow S=80 $$ (assuming $S>0$). In particular
$$fracV_1V_2=frac2555 $$ and so the distance between the bus
and the car is $$80-25-25cdot frac5525=0 $$
EDIT: This solution is not acceptable (and not the intended solution) because if the distance between Alice's and Bob's house is $80$, and they are only moving between each other's houses, they cannot meet $85$ meters from Alice's house. The equation
$$fracV_1V_2=fracS-5555 $$
is correct because the first meeting will necessarily occur before both Alice and Bob have completed their first journey. However, for the second meeting there are other possible cases:
1) It could be that they meet again before Alice has reached Bob's house in her first trip. In this case, the second meeting occurs when in total Alice has travelled $85$ meters and Bob has travelled $S+85$ meters. We obtain
$$ fracV_1V_2=fracS-5555=frac85S+85Rightarrow S^2+30S-2cdot 55cdot 85=0$$
a quadratic equation whose positive root is
$$ S=5(sqrt383-3)approx 82.85$$
Since $82.85<85$, this solution is also unacceptable!
2) The remaining case is the one where they meet again before Bob has reached Alice's house in his first trip. Here at the second meeting in total Bob has travelled $S-85$ meters, while Alice has travelled $2S-85$ meters. We get
$$fracV_1V_2=fracS-5555=frac2S-85S-85 Rightarrow S^2-250S+2cdot 55cdot 85=0$$
a quadratic equation with roots $5(25pm sqrt251)$, of which only
$$S=5(25+sqrt251)approx 204 > 85 $$
is acceptable because $5(25-sqrt251)<85$. The distance between the bus and the car is then
beginalign*x&=S-25-25cdotfracV_2V_1=S-25-25frac55S-55=frac(S-25)(S-55)-25cdot 55S-55=fracS^2-80SS-55=\
&=frac250S-2cdot 55 cdot 85-80SS-55=frac170(S-55)S-55=170
endalign*
This is the intended solution of the problem. Indeed, there are no other possible cases: at the second meeting either Alice or Bob must have completed at least one journey, but none of them can have completed more than one, else there would have been another meeting inbetween.
answered Jul 22 at 13:08
Lorenzo Quarisa
2,284314
Lorenzo Quarisa: in the last row of your solution, how do you substitute $V_2.t_1$ with $$25.frac 5525$$?
– Nhung Huyen
Jul 22 at 19:51
1
OK so the result does not make any sense because if the total distance between the two houses is 80 m, this means that when Alice passed by the bus station (at 25 m from her house), at that moment, Bob was passing by the old car which was (80-25) = 55 m from his house. Therefore until their first meeting (at 55 m away from Bob's house), Bob did not move at all.
– Pradeep Suny
Jul 23 at 7:55
1
I agree that the solution is incorrect, not much because their first meeting is supposed to be at the bus+abandoned car, but rather because if they are only moving between each other's homes and the distance between them is $S=80$, they cannot meet at a distance of $85$ from Alice's house. Therefore the situation OP considered does not lead to any solutions, but there are other cases. I am looking into them now.
– Lorenzo Quarisa
Jul 23 at 8:22
1
@PradeepSuny I have updated my answer with the correct solution (I believe).
– Lorenzo Quarisa
Jul 23 at 9:09
1
Yes this one is correct and the result is 170. I hope the OP also finds it useful.
– Pradeep Suny
Jul 23 at 11:16
 |Â
show 4 more comments
up vote
10
down vote
accepted
You have $V_1t_2=S-55 $, $V_2t_2=55$, $V_1t_3=2S-85$, $V_2t_3=S+85$.
Hence $$fracV_1V_2=fracS-5555=frac2S-85S+85 $$ Thus
$$55(2S-85)=(S+85)(S-55)=S^2+30S-85cdot 55 $$ $$Rightarrow
110S=S^2+30SRightarrow S=80 $$ (assuming $S>0$). In particular
$$fracV_1V_2=frac2555 $$ and so the distance between the bus
and the car is $$80-25-25cdot frac5525=0 $$
EDIT: This solution is not acceptable (and not the intended solution) because if the distance between Alice's and Bob's house is $80$, and they are only moving between each other's houses, they cannot meet $85$ meters from Alice's house. The equation
$$fracV_1V_2=fracS-5555 $$
is correct because the first meeting will necessarily occur before both Alice and Bob have completed their first journey. However, for the second meeting there are other possible cases:
1) It could be that they meet again before Alice has reached Bob's house in her first trip. In this case, the second meeting occurs when in total Alice has travelled $85$ meters and Bob has travelled $S+85$ meters. We obtain
$$ fracV_1V_2=fracS-5555=frac85S+85Rightarrow S^2+30S-2cdot 55cdot 85=0$$
a quadratic equation whose positive root is
$$ S=5(sqrt383-3)approx 82.85$$
Since $82.85<85$, this solution is also unacceptable!
2) The remaining case is the one where they meet again before Bob has reached Alice's house in his first trip. Here at the second meeting in total Bob has travelled $S-85$ meters, while Alice has travelled $2S-85$ meters. We get
$$fracV_1V_2=fracS-5555=frac2S-85S-85 Rightarrow S^2-250S+2cdot 55cdot 85=0$$
a quadratic equation with roots $5(25pm sqrt251)$, of which only
$$S=5(25+sqrt251)approx 204 > 85 $$
is acceptable because $5(25-sqrt251)<85$. The distance between the bus and the car is then
beginalign*x&=S-25-25cdotfracV_2V_1=S-25-25frac55S-55=frac(S-25)(S-55)-25cdot 55S-55=fracS^2-80SS-55=\
&=frac250S-2cdot 55 cdot 85-80SS-55=frac170(S-55)S-55=170
endalign*
This is the intended solution of the problem. Indeed, there are no other possible cases: at the second meeting either Alice or Bob must have completed at least one journey, but none of them can have completed more than one, else there would have been another meeting inbetween.
answered Jul 22 at 13:08
Lorenzo Quarisa
2,284314
Lorenzo Quarisa: in the last row of your solution, how do you substitute $V_2.t_1$ with $$25.frac 5525$$?
– Nhung Huyen
Jul 22 at 19:51
1
OK so the result does not make any sense because if the total distance between the two houses is 80 m, this means that when Alice passed by the bus station (at 25 m from her house), at that moment, Bob was passing by the old car which was (80-25) = 55 m from his house. Therefore until their first meeting (at 55 m away from Bob's house), Bob did not move at all.
– Pradeep Suny
Jul 23 at 7:55
1
I agree that the solution is incorrect, not much because their first meeting is supposed to be at the bus+abandoned car, but rather because if they are only moving between each other's homes and the distance between them is $S=80$, they cannot meet at a distance of $85$ from Alice's house. Therefore the situation OP considered does not lead to any solutions, but there are other cases. I am looking into them now.
– Lorenzo Quarisa
Jul 23 at 8:22
1
@PradeepSuny I have updated my answer with the correct solution (I believe).
– Lorenzo Quarisa
Jul 23 at 9:09
1
Yes this one is correct and the result is 170. I hope the OP also finds it useful.
– Pradeep Suny
Jul 23 at 11:16
 |Â
show 4 more comments
up vote
10
down vote
accepted
up vote
10
down vote
accepted
You have $V_1t_2=S-55 $, $V_2t_2=55$, $V_1t_3=2S-85$, $V_2t_3=S+85$.
Hence $$fracV_1V_2=fracS-5555=frac2S-85S+85 $$ Thus
$$55(2S-85)=(S+85)(S-55)=S^2+30S-85cdot 55 $$ $$Rightarrow
110S=S^2+30SRightarrow S=80 $$ (assuming $S>0$). In particular
$$fracV_1V_2=frac2555 $$ and so the distance between the bus
and the car is $$80-25-25cdot frac5525=0 $$
EDIT: This solution is not acceptable (and not the intended solution) because if the distance between Alice's and Bob's house is $80$, and they are only moving between each other's houses, they cannot meet $85$ meters from Alice's house. The equation
$$fracV_1V_2=fracS-5555 $$
is correct because the first meeting will necessarily occur before both Alice and Bob have completed their first journey. However, for the second meeting there are other possible cases:
1) It could be that they meet again before Alice has reached Bob's house in her first trip. In this case, the second meeting occurs when in total Alice has travelled $85$ meters and Bob has travelled $S+85$ meters. We obtain
$$ fracV_1V_2=fracS-5555=frac85S+85Rightarrow S^2+30S-2cdot 55cdot 85=0$$
a quadratic equation whose positive root is
$$ S=5(sqrt383-3)approx 82.85$$
Since $82.85<85$, this solution is also unacceptable!
2) The remaining case is the one where they meet again before Bob has reached Alice's house in his first trip. Here at the second meeting in total Bob has travelled $S-85$ meters, while Alice has travelled $2S-85$ meters. We get
$$fracV_1V_2=fracS-5555=frac2S-85S-85 Rightarrow S^2-250S+2cdot 55cdot 85=0$$
a quadratic equation with roots $5(25pm sqrt251)$, of which only
$$S=5(25+sqrt251)approx 204 > 85 $$
is acceptable because $5(25-sqrt251)<85$. The distance between the bus and the car is then
beginalign*x&=S-25-25cdotfracV_2V_1=S-25-25frac55S-55=frac(S-25)(S-55)-25cdot 55S-55=fracS^2-80SS-55=\
&=frac250S-2cdot 55 cdot 85-80SS-55=frac170(S-55)S-55=170
endalign*
This is the intended solution of the problem. Indeed, there are no other possible cases: at the second meeting either Alice or Bob must have completed at least one journey, but none of them can have completed more than one, else there would have been another meeting inbetween.
answered Jul 22 at 13:08
Lorenzo Quarisa
2,284314
You have $V_1t_2=S-55 $, $V_2t_2=55$, $V_1t_3=2S-85$, $V_2t_3=S+85$.
Hence $$fracV_1V_2=fracS-5555=frac2S-85S+85 $$ Thus
$$55(2S-85)=(S+85)(S-55)=S^2+30S-85cdot 55 $$ $$Rightarrow
110S=S^2+30SRightarrow S=80 $$ (assuming $S>0$). In particular
$$fracV_1V_2=frac2555 $$ and so the distance between the bus
and the car is $$80-25-25cdot frac5525=0 $$
EDIT: This solution is not acceptable (and not the intended solution) because if the distance between Alice's and Bob's house is $80$, and they are only moving between each other's houses, they cannot meet $85$ meters from Alice's house. The equation
$$fracV_1V_2=fracS-5555 $$
is correct because the first meeting will necessarily occur before both Alice and Bob have completed their first journey. However, for the second meeting there are other possible cases:
1) It could be that they meet again before Alice has reached Bob's house in her first trip. In this case, the second meeting occurs when in total Alice has travelled $85$ meters and Bob has travelled $S+85$ meters. We obtain
$$ fracV_1V_2=fracS-5555=frac85S+85Rightarrow S^2+30S-2cdot 55cdot 85=0$$
a quadratic equation whose positive root is
$$ S=5(sqrt383-3)approx 82.85$$
Since $82.85<85$, this solution is also unacceptable!
2) The remaining case is the one where they meet again before Bob has reached Alice's house in his first trip. Here at the second meeting in total Bob has travelled $S-85$ meters, while Alice has travelled $2S-85$ meters. We get
$$fracV_1V_2=fracS-5555=frac2S-85S-85 Rightarrow S^2-250S+2cdot 55cdot 85=0$$
a quadratic equation with roots $5(25pm sqrt251)$, of which only
$$S=5(25+sqrt251)approx 204 > 85 $$
is acceptable because $5(25-sqrt251)<85$. The distance between the bus and the car is then
beginalign*x&=S-25-25cdotfracV_2V_1=S-25-25frac55S-55=frac(S-25)(S-55)-25cdot 55S-55=fracS^2-80SS-55=\
&=frac250S-2cdot 55 cdot 85-80SS-55=frac170(S-55)S-55=170
endalign*
This is the intended solution of the problem. Indeed, there are no other possible cases: at the second meeting either Alice or Bob must have completed at least one journey, but none of them can have completed more than one, else there would have been another meeting inbetween.
answered Jul 22 at 13:08
Lorenzo Quarisa
2,284314
edited Jul 23 at 11:43
answered Jul 22 at 13:08
Lorenzo Quarisa
2,284314
answered Jul 22 at 13:08
Lorenzo Quarisa
2,284314
answered Jul 22 at 13:08
Lorenzo Quarisa
2,284314
2,284314
Lorenzo Quarisa: in the last row of your solution, how do you substitute $V_2.t_1$ with $$25.frac 5525$$?
– Nhung Huyen
Jul 22 at 19:51
1
OK so the result does not make any sense because if the total distance between the two houses is 80 m, this means that when Alice passed by the bus station (at 25 m from her house), at that moment, Bob was passing by the old car which was (80-25) = 55 m from his house. Therefore until their first meeting (at 55 m away from Bob's house), Bob did not move at all.
– Pradeep Suny
Jul 23 at 7:55
1
I agree that the solution is incorrect, not much because their first meeting is supposed to be at the bus+abandoned car, but rather because if they are only moving between each other's homes and the distance between them is $S=80$, they cannot meet at a distance of $85$ from Alice's house. Therefore the situation OP considered does not lead to any solutions, but there are other cases. I am looking into them now.
– Lorenzo Quarisa
Jul 23 at 8:22
1
@PradeepSuny I have updated my answer with the correct solution (I believe).
– Lorenzo Quarisa
Jul 23 at 9:09
1
Yes this one is correct and the result is 170. I hope the OP also finds it useful.
– Pradeep Suny
Jul 23 at 11:16
 |Â
show 4 more comments
Lorenzo Quarisa: in the last row of your solution, how do you substitute $V_2.t_1$ with $$25.frac 5525$$?
– Nhung Huyen
Jul 22 at 19:51
1
OK so the result does not make any sense because if the total distance between the two houses is 80 m, this means that when Alice passed by the bus station (at 25 m from her house), at that moment, Bob was passing by the old car which was (80-25) = 55 m from his house. Therefore until their first meeting (at 55 m away from Bob's house), Bob did not move at all.
– Pradeep Suny
Jul 23 at 7:55
1
I agree that the solution is incorrect, not much because their first meeting is supposed to be at the bus+abandoned car, but rather because if they are only moving between each other's homes and the distance between them is $S=80$, they cannot meet at a distance of $85$ from Alice's house. Therefore the situation OP considered does not lead to any solutions, but there are other cases. I am looking into them now.
– Lorenzo Quarisa
Jul 23 at 8:22
1
@PradeepSuny I have updated my answer with the correct solution (I believe).
– Lorenzo Quarisa
Jul 23 at 9:09
1
Yes this one is correct and the result is 170. I hope the OP also finds it useful.
– Pradeep Suny
Jul 23 at 11:16
Lorenzo Quarisa: in the last row of your solution, how do you substitute $V_2.t_1$ with $$25.frac 5525$$?
– Nhung Huyen
Jul 22 at 19:51
Lorenzo Quarisa: in the last row of your solution, how do you substitute $V_2.t_1$ with $$25.frac 5525$$?
– Nhung Huyen
Jul 22 at 19:51
1
1
OK so the result does not make any sense because if the total distance between the two houses is 80 m, this means that when Alice passed by the bus station (at 25 m from her house), at that moment, Bob was passing by the old car which was (80-25) = 55 m from his house. Therefore until their first meeting (at 55 m away from Bob's house), Bob did not move at all.
– Pradeep Suny
Jul 23 at 7:55
OK so the result does not make any sense because if the total distance between the two houses is 80 m, this means that when Alice passed by the bus station (at 25 m from her house), at that moment, Bob was passing by the old car which was (80-25) = 55 m from his house. Therefore until their first meeting (at 55 m away from Bob's house), Bob did not move at all.
– Pradeep Suny
Jul 23 at 7:55
1
1
I agree that the solution is incorrect, not much because their first meeting is supposed to be at the bus+abandoned car, but rather because if they are only moving between each other's homes and the distance between them is $S=80$, they cannot meet at a distance of $85$ from Alice's house. Therefore the situation OP considered does not lead to any solutions, but there are other cases. I am looking into them now.
– Lorenzo Quarisa
Jul 23 at 8:22
I agree that the solution is incorrect, not much because their first meeting is supposed to be at the bus+abandoned car, but rather because if they are only moving between each other's homes and the distance between them is $S=80$, they cannot meet at a distance of $85$ from Alice's house. Therefore the situation OP considered does not lead to any solutions, but there are other cases. I am looking into them now.
– Lorenzo Quarisa
Jul 23 at 8:22
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@PradeepSuny I have updated my answer with the correct solution (I believe).
– Lorenzo Quarisa
Jul 23 at 9:09
@PradeepSuny I have updated my answer with the correct solution (I believe).
– Lorenzo Quarisa
Jul 23 at 9:09
1
1
Yes this one is correct and the result is 170. I hope the OP also finds it useful.
– Pradeep Suny
Jul 23 at 11:16
Yes this one is correct and the result is 170. I hope the OP also finds it useful.
– Pradeep Suny
Jul 23 at 11:16
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I'm wondering if it's possible to solve this problem in a geometrical way, e.g. with a time distance diagram.
– Eric Duminil
Jul 22 at 20:25