When Kobie goes to school, he walks half the time and runs half the time. [closed]
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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 Nov 19 at 22:23
Andrew Li
4,1421927
asked Nov 19 at 22:23
Isaiah Leobrera
301
closed as off-topic by Leucippus, José Carlos Santos, Shailesh, amWhy, user10354138 Nov 20 at 1:03
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 Nov 19 at 22:23
Andrew Li
4,1421927
asked Nov 19 at 22:23
Isaiah Leobrera
301
closed as off-topic by Leucippus, José Carlos Santos, Shailesh, amWhy, user10354138 Nov 20 at 1:03
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
Nov 19 at 22:36
add a comment |
up vote
0
down vote
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 Nov 19 at 22:23
Andrew Li
4,1421927
asked Nov 19 at 22:23
Isaiah Leobrera
301
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 Nov 19 at 22:23
Andrew Li
4,1421927
asked Nov 19 at 22:23
Isaiah Leobrera
301
edited Nov 19 at 22:23
Andrew Li
4,1421927
asked Nov 19 at 22:23
Isaiah Leobrera
301
edited Nov 19 at 22:23
Andrew Li
4,1421927
edited Nov 19 at 22:23
Andrew Li
4,1421927
edited Nov 19 at 22:23
Andrew Li
4,1421927
4,1421927
asked Nov 19 at 22:23
Isaiah Leobrera
301
asked Nov 19 at 22:23
Isaiah Leobrera
301
asked Nov 19 at 22:23
Isaiah Leobrera
301
301
closed as off-topic by Leucippus, José Carlos Santos, Shailesh, amWhy, user10354138 Nov 20 at 1:03
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.
closed as off-topic by Leucippus, José Carlos Santos, Shailesh, amWhy, user10354138 Nov 20 at 1:03
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
Nov 19 at 22:36
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
Nov 19 at 22:36
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
Nov 19 at 22:36
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
Nov 19 at 22:36
add a comment |
2 Answers
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up vote
2
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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 Nov 19 at 22:35
Andrei
10.7k21025
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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 Nov 19 at 22:41
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
Nov 19 at 22:49
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 Nov 19 at 22:35
Andrei
10.7k21025
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 Nov 19 at 22:35
Andrei
10.7k21025
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 Nov 19 at 22:35
Andrei
10.7k21025
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 Nov 19 at 22:35
Andrei
10.7k21025
edited Nov 19 at 22:44
answered Nov 19 at 22:35
Andrei
10.7k21025
answered Nov 19 at 22:35
Andrei
10.7k21025
answered Nov 19 at 22:35
Andrei
10.7k21025
10.7k21025
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 Nov 19 at 22:41
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
Nov 19 at 22:49
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 Nov 19 at 22:41
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
Nov 19 at 22:49
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 Nov 19 at 22:41
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 Nov 19 at 22:41
uniquesolution
8,631823
answered Nov 19 at 22:41
uniquesolution
8,631823
answered Nov 19 at 22:41
uniquesolution
8,631823
answered Nov 19 at 22:41
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
Nov 19 at 22:49
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
Nov 19 at 22:49
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
Nov 19 at 22:49
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
Nov 19 at 22:49
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
Nov 19 at 22:36