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When Kobie goes to school, he walks half the time and runs half the time. [on hold]
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When Kobie goes to school, he walks half the time and runs half the time. When he comes
home from school, he walks half the distance and runs half the distance. If he runs twice as
fast as he walks, find the ratio of the time it takes for him to get to school, to the time it takes
for him to come home from school.
I tried messing around with the distance formula and got that the walking speed is 2/3 (not sure if this is correct), but
I am not sure how to go on frome here.
algebra-precalculus
edited 2 days ago
Andrew Li
4,1251927
asked 2 days ago
Isaiah Leobrera
221
put on hold as off-topic by Leucippus, José Carlos Santos, Shailesh, amWhy, user10354138 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, José Carlos Santos, Shailesh, amWhy, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
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When Kobie goes to school, he walks half the time and runs half the time. When he comes
home from school, he walks half the distance and runs half the distance. If he runs twice as
fast as he walks, find the ratio of the time it takes for him to get to school, to the time it takes
for him to come home from school.
I tried messing around with the distance formula and got that the walking speed is 2/3 (not sure if this is correct), but
I am not sure how to go on frome here.
algebra-precalculus
edited 2 days ago
Andrew Li
4,1251927
asked 2 days ago
Isaiah Leobrera
221
put on hold as off-topic by Leucippus, José Carlos Santos, Shailesh, amWhy, user10354138 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, José Carlos Santos, Shailesh, amWhy, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
1
2/3 of what? The best things you can do to start a problem like this are (1) set out all the starting conditions (e.g. "let his distance to school = d, let walking speed = s", etc) and (2) draw a diagram to better understand what's happening.
– ConMan
2 days ago
add a comment |
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0
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favorite
up vote
0
down vote
favorite
When Kobie goes to school, he walks half the time and runs half the time. When he comes
home from school, he walks half the distance and runs half the distance. If he runs twice as
fast as he walks, find the ratio of the time it takes for him to get to school, to the time it takes
for him to come home from school.
I tried messing around with the distance formula and got that the walking speed is 2/3 (not sure if this is correct), but
I am not sure how to go on frome here.
algebra-precalculus
edited 2 days ago
Andrew Li
4,1251927
asked 2 days ago
Isaiah Leobrera
221
When Kobie goes to school, he walks half the time and runs half the time. When he comes
home from school, he walks half the distance and runs half the distance. If he runs twice as
fast as he walks, find the ratio of the time it takes for him to get to school, to the time it takes
for him to come home from school.
I tried messing around with the distance formula and got that the walking speed is 2/3 (not sure if this is correct), but
I am not sure how to go on frome here.
algebra-precalculus
algebra-precalculus
edited 2 days ago
Andrew Li
4,1251927
asked 2 days ago
Isaiah Leobrera
221
edited 2 days ago
Andrew Li
4,1251927
asked 2 days ago
Isaiah Leobrera
221
edited 2 days ago
Andrew Li
4,1251927
edited 2 days ago
Andrew Li
4,1251927
edited 2 days ago
Andrew Li
4,1251927
4,1251927
asked 2 days ago
Isaiah Leobrera
221
asked 2 days ago
Isaiah Leobrera
221
asked 2 days ago
Isaiah Leobrera
221
221
put on hold as off-topic by Leucippus, José Carlos Santos, Shailesh, amWhy, user10354138 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, José Carlos Santos, Shailesh, amWhy, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Leucippus, José Carlos Santos, Shailesh, amWhy, user10354138 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, José Carlos Santos, Shailesh, amWhy, user10354138
If this question can be reworded to fit the rules in the help center, please edit the question.
1
2/3 of what? The best things you can do to start a problem like this are (1) set out all the starting conditions (e.g. "let his distance to school = d, let walking speed = s", etc) and (2) draw a diagram to better understand what's happening.
– ConMan
2 days ago
add a comment |
1
2/3 of what? The best things you can do to start a problem like this are (1) set out all the starting conditions (e.g. "let his distance to school = d, let walking speed = s", etc) and (2) draw a diagram to better understand what's happening.
– ConMan
2 days ago
1
1
2/3 of what? The best things you can do to start a problem like this are (1) set out all the starting conditions (e.g. "let his distance to school = d, let walking speed = s", etc) and (2) draw a diagram to better understand what's happening.
– ConMan
2 days ago
2/3 of what? The best things you can do to start a problem like this are (1) set out all the starting conditions (e.g. "let his distance to school = d, let walking speed = s", etc) and (2) draw a diagram to better understand what's happening.
– ConMan
2 days ago
add a comment |
2 Answers
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When he goes to school, he takes time $t_1$ to travel a distance $d$. We can write this distance in terms of velocities, $v_w$ and $v_r$ (for walking and running):
$$d=v_wfrac{t_1}2+v_rfrac{t_1}2=frac{v_w+v_r}2t_1$$
We now use a similar approach to calculate the time $t_2$ it takes to return from school. It takes $frac d{2}frac1{v_w}$ to walk, and $frac d{2}frac1{v_r}$ to run, so $$t_2= frac d{2}frac1{v_w}+frac d{2}frac1{v_r}$$
From these two equations you can find the ratio. Just plug in $d$ from the first into the second.
answered 2 days ago
Andrei
9,82621024
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The ratio is $4/9$.
Kobie's speed of walking, is, say, $v$ km per hour. So his running speed is $2v$ km per hour. Suppose the distance from Kobie's home to school is $d$ km. Assume it takes him $T$ hours to get to school. Let's say that $d_1$ is the distance Kobie covers while walking on his way to school, and $d_2$ is the distance Kobie covers while running on his way to school. Then $d_1+d_2=d$, and
$$T/2=d_1/v=d_2/(2v)$$
hence $d_1=d_2/2$, so $d=3d_1$.
On his way back from school, the time it takes him to cover half the distance while running is $(d/2)/(v/2)=d/v$, and the time it takes him to the other half of the distance while walking, is $(d/2)/v$. Hence the ratio of times is
$$frac{2d_1/v}{(d/v+(d/(2v))}=frac{2d_1}{d+d/2}=frac{2d_1}{3d_1+3d_1/2}=frac{2}{3+3/2}=frac{2}{9/2}=frac{4}{9}$$
answered 2 days ago
uniquesolution
8,631823
Small error: it seems like you use more time to run than to walk the half distance. The velocity for running is $2v$ not $v/2$, so the time for that segment is $d/(4v)$ not $d/v$. The final answer is $8/9$
– Andrei
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
When he goes to school, he takes time $t_1$ to travel a distance $d$. We can write this distance in terms of velocities, $v_w$ and $v_r$ (for walking and running):
$$d=v_wfrac{t_1}2+v_rfrac{t_1}2=frac{v_w+v_r}2t_1$$
We now use a similar approach to calculate the time $t_2$ it takes to return from school. It takes $frac d{2}frac1{v_w}$ to walk, and $frac d{2}frac1{v_r}$ to run, so $$t_2= frac d{2}frac1{v_w}+frac d{2}frac1{v_r}$$
From these two equations you can find the ratio. Just plug in $d$ from the first into the second.
answered 2 days ago
Andrei
9,82621024
add a comment |
up vote
2
down vote
When he goes to school, he takes time $t_1$ to travel a distance $d$. We can write this distance in terms of velocities, $v_w$ and $v_r$ (for walking and running):
$$d=v_wfrac{t_1}2+v_rfrac{t_1}2=frac{v_w+v_r}2t_1$$
We now use a similar approach to calculate the time $t_2$ it takes to return from school. It takes $frac d{2}frac1{v_w}$ to walk, and $frac d{2}frac1{v_r}$ to run, so $$t_2= frac d{2}frac1{v_w}+frac d{2}frac1{v_r}$$
From these two equations you can find the ratio. Just plug in $d$ from the first into the second.
answered 2 days ago
Andrei
9,82621024
add a comment |
up vote
2
down vote
up vote
2
down vote
When he goes to school, he takes time $t_1$ to travel a distance $d$. We can write this distance in terms of velocities, $v_w$ and $v_r$ (for walking and running):
$$d=v_wfrac{t_1}2+v_rfrac{t_1}2=frac{v_w+v_r}2t_1$$
We now use a similar approach to calculate the time $t_2$ it takes to return from school. It takes $frac d{2}frac1{v_w}$ to walk, and $frac d{2}frac1{v_r}$ to run, so $$t_2= frac d{2}frac1{v_w}+frac d{2}frac1{v_r}$$
From these two equations you can find the ratio. Just plug in $d$ from the first into the second.
answered 2 days ago
Andrei
9,82621024
When he goes to school, he takes time $t_1$ to travel a distance $d$. We can write this distance in terms of velocities, $v_w$ and $v_r$ (for walking and running):
$$d=v_wfrac{t_1}2+v_rfrac{t_1}2=frac{v_w+v_r}2t_1$$
We now use a similar approach to calculate the time $t_2$ it takes to return from school. It takes $frac d{2}frac1{v_w}$ to walk, and $frac d{2}frac1{v_r}$ to run, so $$t_2= frac d{2}frac1{v_w}+frac d{2}frac1{v_r}$$
From these two equations you can find the ratio. Just plug in $d$ from the first into the second.
answered 2 days ago
Andrei
9,82621024
edited 2 days ago
answered 2 days ago
Andrei
9,82621024
answered 2 days ago
Andrei
9,82621024
answered 2 days ago
Andrei
9,82621024
9,82621024
add a comment |
add a comment |
up vote
1
down vote
The ratio is $4/9$.
Kobie's speed of walking, is, say, $v$ km per hour. So his running speed is $2v$ km per hour. Suppose the distance from Kobie's home to school is $d$ km. Assume it takes him $T$ hours to get to school. Let's say that $d_1$ is the distance Kobie covers while walking on his way to school, and $d_2$ is the distance Kobie covers while running on his way to school. Then $d_1+d_2=d$, and
$$T/2=d_1/v=d_2/(2v)$$
hence $d_1=d_2/2$, so $d=3d_1$.
On his way back from school, the time it takes him to cover half the distance while running is $(d/2)/(v/2)=d/v$, and the time it takes him to the other half of the distance while walking, is $(d/2)/v$. Hence the ratio of times is
$$frac{2d_1/v}{(d/v+(d/(2v))}=frac{2d_1}{d+d/2}=frac{2d_1}{3d_1+3d_1/2}=frac{2}{3+3/2}=frac{2}{9/2}=frac{4}{9}$$
answered 2 days ago
uniquesolution
8,631823
Small error: it seems like you use more time to run than to walk the half distance. The velocity for running is $2v$ not $v/2$, so the time for that segment is $d/(4v)$ not $d/v$. The final answer is $8/9$
– Andrei
2 days ago
add a comment |
up vote
1
down vote
The ratio is $4/9$.
Kobie's speed of walking, is, say, $v$ km per hour. So his running speed is $2v$ km per hour. Suppose the distance from Kobie's home to school is $d$ km. Assume it takes him $T$ hours to get to school. Let's say that $d_1$ is the distance Kobie covers while walking on his way to school, and $d_2$ is the distance Kobie covers while running on his way to school. Then $d_1+d_2=d$, and
$$T/2=d_1/v=d_2/(2v)$$
hence $d_1=d_2/2$, so $d=3d_1$.
On his way back from school, the time it takes him to cover half the distance while running is $(d/2)/(v/2)=d/v$, and the time it takes him to the other half of the distance while walking, is $(d/2)/v$. Hence the ratio of times is
$$frac{2d_1/v}{(d/v+(d/(2v))}=frac{2d_1}{d+d/2}=frac{2d_1}{3d_1+3d_1/2}=frac{2}{3+3/2}=frac{2}{9/2}=frac{4}{9}$$
answered 2 days ago
uniquesolution
8,631823
Small error: it seems like you use more time to run than to walk the half distance. The velocity for running is $2v$ not $v/2$, so the time for that segment is $d/(4v)$ not $d/v$. The final answer is $8/9$
– Andrei
2 days ago
add a comment |
up vote
1
down vote
up vote
1
down vote
The ratio is $4/9$.
Kobie's speed of walking, is, say, $v$ km per hour. So his running speed is $2v$ km per hour. Suppose the distance from Kobie's home to school is $d$ km. Assume it takes him $T$ hours to get to school. Let's say that $d_1$ is the distance Kobie covers while walking on his way to school, and $d_2$ is the distance Kobie covers while running on his way to school. Then $d_1+d_2=d$, and
$$T/2=d_1/v=d_2/(2v)$$
hence $d_1=d_2/2$, so $d=3d_1$.
On his way back from school, the time it takes him to cover half the distance while running is $(d/2)/(v/2)=d/v$, and the time it takes him to the other half of the distance while walking, is $(d/2)/v$. Hence the ratio of times is
$$frac{2d_1/v}{(d/v+(d/(2v))}=frac{2d_1}{d+d/2}=frac{2d_1}{3d_1+3d_1/2}=frac{2}{3+3/2}=frac{2}{9/2}=frac{4}{9}$$
answered 2 days ago
uniquesolution
8,631823
The ratio is $4/9$.
Kobie's speed of walking, is, say, $v$ km per hour. So his running speed is $2v$ km per hour. Suppose the distance from Kobie's home to school is $d$ km. Assume it takes him $T$ hours to get to school. Let's say that $d_1$ is the distance Kobie covers while walking on his way to school, and $d_2$ is the distance Kobie covers while running on his way to school. Then $d_1+d_2=d$, and
$$T/2=d_1/v=d_2/(2v)$$
hence $d_1=d_2/2$, so $d=3d_1$.
On his way back from school, the time it takes him to cover half the distance while running is $(d/2)/(v/2)=d/v$, and the time it takes him to the other half of the distance while walking, is $(d/2)/v$. Hence the ratio of times is
$$frac{2d_1/v}{(d/v+(d/(2v))}=frac{2d_1}{d+d/2}=frac{2d_1}{3d_1+3d_1/2}=frac{2}{3+3/2}=frac{2}{9/2}=frac{4}{9}$$
answered 2 days ago
uniquesolution
8,631823
answered 2 days ago
uniquesolution
8,631823
answered 2 days ago
uniquesolution
8,631823
answered 2 days ago
uniquesolution
8,631823
8,631823
Small error: it seems like you use more time to run than to walk the half distance. The velocity for running is $2v$ not $v/2$, so the time for that segment is $d/(4v)$ not $d/v$. The final answer is $8/9$
– Andrei
2 days ago
add a comment |
Small error: it seems like you use more time to run than to walk the half distance. The velocity for running is $2v$ not $v/2$, so the time for that segment is $d/(4v)$ not $d/v$. The final answer is $8/9$
– Andrei
2 days ago
Small error: it seems like you use more time to run than to walk the half distance. The velocity for running is $2v$ not $v/2$, so the time for that segment is $d/(4v)$ not $d/v$. The final answer is $8/9$
– Andrei
2 days ago
Small error: it seems like you use more time to run than to walk the half distance. The velocity for running is $2v$ not $v/2$, so the time for that segment is $d/(4v)$ not $d/v$. The final answer is $8/9$
– Andrei
2 days ago
add a comment |
1
2/3 of what? The best things you can do to start a problem like this are (1) set out all the starting conditions (e.g. "let his distance to school = d, let walking speed = s", etc) and (2) draw a diagram to better understand what's happening.
– ConMan
2 days ago