Finding average speed of a driver

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I was asked this question by a friend.

Ann leaves Town A and drives towards Town B while Leslie simultaneously leaves Town B and drives towards Town A. They travel on the same route and, after passing each other on the way, Ann takes 4 hours to reach her destination, while Leslie takes 9 hours to reach her destination. If the average speed of Ann's car is 48 miles per hour, what is the average speed of Leslie's car, in miles per hour?

I am not good with these types of questions and I tried my best. I tried to use $d=rt$ but couldn't incorporate the times for different portions of the trip together.

How can I go about solving this problem?

asked Jul 25 at 22:57

silverpegasus258

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up vote
0
down vote

favorite

I was asked this question by a friend.

Ann leaves Town A and drives towards Town B while Leslie simultaneously leaves Town B and drives towards Town A. They travel on the same route and, after passing each other on the way, Ann takes 4 hours to reach her destination, while Leslie takes 9 hours to reach her destination. If the average speed of Ann's car is 48 miles per hour, what is the average speed of Leslie's car, in miles per hour?

I am not good with these types of questions and I tried my best. I tried to use $d=rt$ but couldn't incorporate the times for different portions of the trip together.

How can I go about solving this problem?

asked Jul 25 at 22:57

silverpegasus258

41829

add a commentÂ |Â

up vote
0
down vote

favorite

I was asked this question by a friend.

Ann leaves Town A and drives towards Town B while Leslie simultaneously leaves Town B and drives towards Town A. They travel on the same route and, after passing each other on the way, Ann takes 4 hours to reach her destination, while Leslie takes 9 hours to reach her destination. If the average speed of Ann's car is 48 miles per hour, what is the average speed of Leslie's car, in miles per hour?

I am not good with these types of questions and I tried my best. I tried to use $d=rt$ but couldn't incorporate the times for different portions of the trip together.

How can I go about solving this problem?

asked Jul 25 at 22:57

silverpegasus258

41829

I was asked this question by a friend.

Ann leaves Town A and drives towards Town B while Leslie simultaneously leaves Town B and drives towards Town A. They travel on the same route and, after passing each other on the way, Ann takes 4 hours to reach her destination, while Leslie takes 9 hours to reach her destination. If the average speed of Ann's car is 48 miles per hour, what is the average speed of Leslie's car, in miles per hour?

I am not good with these types of questions and I tried my best. I tried to use $d=rt$ but couldn't incorporate the times for different portions of the trip together.

How can I go about solving this problem?

asked Jul 25 at 22:57

silverpegasus258

41829

asked Jul 25 at 22:57

silverpegasus258

41829

asked Jul 25 at 22:57

silverpegasus258

41829

asked Jul 25 at 22:57

silverpegasus258

41829

add a commentÂ |Â

5 Answers
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The ratio of times for both people must be the same for the sections of the journey. If we assign t to the time it took for both Ann and Leslie to complete the first part, then:

$frac4t = fract9$

$t^2 = 36$

$t = 6$ hours

So Leslie took 6 hours to travel $4cdot 48 = 192$ miles

Leslie's speed is therefore $frac1926 = 32$ mph

answered Jul 26 at 0:19

Phil H

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So we have that both travel the same distance. Let $v_A$ be Anna's average speed and $v_L$ Lesslie's average speed. Let's denote the distance with $s$ and the whole time taken for Lesslie to reach her destination be $t$ then we have:
$$tv_L = s$$
$$(t-5)v_A = s$$
and for the time taken for them to meet each other we have:
$$ (t-9)(v_A + v_L) = s$$
Solve the system of equations.

answered Jul 25 at 23:06

user1949350

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Let Ann's position be $d_A$, and Leslie's position be $d_B$. Further, let's measure distance relative to town A, and say that the distance to town B is $D$. Since they're travelling at different, but constant, speeds, we can write the following:

$d_A = v_A t \
d_L = D - v_L t$

Notice that I've arranged Leslie's position so that at $t = 0$ she's in town B.

At some time $t = t_0$, $d_A = d_L$ (i.e. their positions are the same) - in other words, $v_A t_0 = D - v_L t_0$. Then, 4 hours later Ann reaches town B, meaning:

$v_A (t_0 + 4) = D$

And 9 hours later, Leslie reaches town A, meaning:

$D - v_L (t_0 + 9) = 0$

So we have three equations to work with:

$(v_A + v_L) t_0 = D \
v_A (t_0 + 4) = D \
v_L (t_0 + 9) = D$

You can eliminate $D$ and $t_0$ across the equations, letting you express $v_L$ in terms of $v_A$, which we've been told is 48.

answered Jul 25 at 23:24

ConMan

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up vote
0
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A way to consider this type of problem while you are still grappling with the basics of algebra is to just consider it logically as follows:

You know how long it takes Leslie to complete her journey, so if you knew how far her journey was, you would have an answer immediately. Averagespeed = distance/time

You know how long it took Ann to complete her journey, and you know her average speed, so you can calculate the distance of her journey. Distance = Averagespeed*time

You know the two journeys were over the same distance as A to B distance must equal B to A distance. So now, knowing Ann's journey distance, you know Leslie's journey distance.

So finally, you have both distance and time for Leslie and can immediately calculate the average speed of her journey (as shown above).

edited Jul 25 at 23:28

answered Jul 25 at 23:27

Penguino

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Leslie drive at speed $v_1.$ Anne drives at speed $v_2 = 48 frac mih$

Suppose $t$ hours pass from the time of departure to the time of meeting.

In that time Anne covers $v_2t$ miles and Leslie covers $v_1t$ miles.

over the rest of the journey each covers the distance that the other traveled and the times are given.

$frac v_1tv_2 = 4$ and $frac v_2tv_1 = 9$

Yow we can cross cross multiply each plug $v_2 = 48$ and sustitue from one equation to the other.

But I think would be more fun to divide one by the other.

$dfrac frac v_1tv_2frac v_2tv_1 = frac 49\
frac v_1^2v_2^2 = frac 49\
v_1 = frac 23 v_2$

answered Jul 25 at 23:34

Doug M

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5 Answers
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active

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5 Answers
5

active

oldest

votes

up vote
1
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The ratio of times for both people must be the same for the sections of the journey. If we assign t to the time it took for both Ann and Leslie to complete the first part, then:

$frac4t = fract9$

$t^2 = 36$

$t = 6$ hours

So Leslie took 6 hours to travel $4cdot 48 = 192$ miles

Leslie's speed is therefore $frac1926 = 32$ mph

answered Jul 26 at 0:19

Phil H

1,8232311

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up vote
1
down vote

The ratio of times for both people must be the same for the sections of the journey. If we assign t to the time it took for both Ann and Leslie to complete the first part, then:

$frac4t = fract9$

$t^2 = 36$

$t = 6$ hours

So Leslie took 6 hours to travel $4cdot 48 = 192$ miles

Leslie's speed is therefore $frac1926 = 32$ mph

answered Jul 26 at 0:19

Phil H

1,8232311

add a commentÂ |Â

up vote
1
down vote

The ratio of times for both people must be the same for the sections of the journey. If we assign t to the time it took for both Ann and Leslie to complete the first part, then:

$frac4t = fract9$

$t^2 = 36$

$t = 6$ hours

So Leslie took 6 hours to travel $4cdot 48 = 192$ miles

Leslie's speed is therefore $frac1926 = 32$ mph

answered Jul 26 at 0:19

Phil H

1,8232311

The ratio of times for both people must be the same for the sections of the journey. If we assign t to the time it took for both Ann and Leslie to complete the first part, then:

$frac4t = fract9$

$t^2 = 36$

$t = 6$ hours

So Leslie took 6 hours to travel $4cdot 48 = 192$ miles

Leslie's speed is therefore $frac1926 = 32$ mph

answered Jul 26 at 0:19

Phil H

1,8232311

answered Jul 26 at 0:19

Phil H

1,8232311

answered Jul 26 at 0:19

Phil H

1,8232311

answered Jul 26 at 0:19

Phil H

1,8232311

add a commentÂ |Â

up vote
0
down vote

answered Jul 25 at 23:06

user1949350

977

add a commentÂ |Â

up vote
0
down vote

answered Jul 25 at 23:06

user1949350

977

add a commentÂ |Â

up vote
0
down vote

answered Jul 25 at 23:06

user1949350

977

answered Jul 25 at 23:06

user1949350

977

answered Jul 25 at 23:06

user1949350

977

answered Jul 25 at 23:06

user1949350

977

answered Jul 25 at 23:06

user1949350

977

add a commentÂ |Â

up vote
0
down vote

$d_A = v_A t \
d_L = D - v_L t$

Notice that I've arranged Leslie's position so that at $t = 0$ she's in town B.

At some time $t = t_0$, $d_A = d_L$ (i.e. their positions are the same) - in other words, $v_A t_0 = D - v_L t_0$. Then, 4 hours later Ann reaches town B, meaning:

$v_A (t_0 + 4) = D$

And 9 hours later, Leslie reaches town A, meaning:

$D - v_L (t_0 + 9) = 0$

So we have three equations to work with:

$(v_A + v_L) t_0 = D \
v_A (t_0 + 4) = D \
v_L (t_0 + 9) = D$

You can eliminate $D$ and $t_0$ across the equations, letting you express $v_L$ in terms of $v_A$, which we've been told is 48.

answered Jul 25 at 23:24

ConMan

6,9351324

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up vote
0
down vote

$d_A = v_A t \
d_L = D - v_L t$

Notice that I've arranged Leslie's position so that at $t = 0$ she's in town B.

At some time $t = t_0$, $d_A = d_L$ (i.e. their positions are the same) - in other words, $v_A t_0 = D - v_L t_0$. Then, 4 hours later Ann reaches town B, meaning:

$v_A (t_0 + 4) = D$

And 9 hours later, Leslie reaches town A, meaning:

$D - v_L (t_0 + 9) = 0$

So we have three equations to work with:

$(v_A + v_L) t_0 = D \
v_A (t_0 + 4) = D \
v_L (t_0 + 9) = D$

You can eliminate $D$ and $t_0$ across the equations, letting you express $v_L$ in terms of $v_A$, which we've been told is 48.

answered Jul 25 at 23:24

ConMan

6,9351324

add a commentÂ |Â

up vote
0
down vote

$d_A = v_A t \
d_L = D - v_L t$

Notice that I've arranged Leslie's position so that at $t = 0$ she's in town B.

At some time $t = t_0$, $d_A = d_L$ (i.e. their positions are the same) - in other words, $v_A t_0 = D - v_L t_0$. Then, 4 hours later Ann reaches town B, meaning:

$v_A (t_0 + 4) = D$

And 9 hours later, Leslie reaches town A, meaning:

$D - v_L (t_0 + 9) = 0$

So we have three equations to work with:

$(v_A + v_L) t_0 = D \
v_A (t_0 + 4) = D \
v_L (t_0 + 9) = D$

You can eliminate $D$ and $t_0$ across the equations, letting you express $v_L$ in terms of $v_A$, which we've been told is 48.

answered Jul 25 at 23:24

ConMan

6,9351324

$d_A = v_A t \
d_L = D - v_L t$

Notice that I've arranged Leslie's position so that at $t = 0$ she's in town B.

At some time $t = t_0$, $d_A = d_L$ (i.e. their positions are the same) - in other words, $v_A t_0 = D - v_L t_0$. Then, 4 hours later Ann reaches town B, meaning:

$v_A (t_0 + 4) = D$

And 9 hours later, Leslie reaches town A, meaning:

$D - v_L (t_0 + 9) = 0$

So we have three equations to work with:

$(v_A + v_L) t_0 = D \
v_A (t_0 + 4) = D \
v_L (t_0 + 9) = D$

You can eliminate $D$ and $t_0$ across the equations, letting you express $v_L$ in terms of $v_A$, which we've been told is 48.

answered Jul 25 at 23:24

ConMan

6,9351324

answered Jul 25 at 23:24

ConMan

6,9351324

answered Jul 25 at 23:24

ConMan

6,9351324

answered Jul 25 at 23:24

ConMan

6,9351324

add a commentÂ |Â

up vote
0
down vote

A way to consider this type of problem while you are still grappling with the basics of algebra is to just consider it logically as follows:

You know how long it takes Leslie to complete her journey, so if you knew how far her journey was, you would have an answer immediately. Averagespeed = distance/time

You know how long it took Ann to complete her journey, and you know her average speed, so you can calculate the distance of her journey. Distance = Averagespeed*time

You know the two journeys were over the same distance as A to B distance must equal B to A distance. So now, knowing Ann's journey distance, you know Leslie's journey distance.

So finally, you have both distance and time for Leslie and can immediately calculate the average speed of her journey (as shown above).

edited Jul 25 at 23:28

answered Jul 25 at 23:27

Penguino

74449

add a commentÂ |Â

up vote
0
down vote

A way to consider this type of problem while you are still grappling with the basics of algebra is to just consider it logically as follows:

You know how long it takes Leslie to complete her journey, so if you knew how far her journey was, you would have an answer immediately. Averagespeed = distance/time

You know how long it took Ann to complete her journey, and you know her average speed, so you can calculate the distance of her journey. Distance = Averagespeed*time

You know the two journeys were over the same distance as A to B distance must equal B to A distance. So now, knowing Ann's journey distance, you know Leslie's journey distance.

So finally, you have both distance and time for Leslie and can immediately calculate the average speed of her journey (as shown above).

edited Jul 25 at 23:28

answered Jul 25 at 23:27

Penguino

74449

add a commentÂ |Â

up vote
0
down vote

A way to consider this type of problem while you are still grappling with the basics of algebra is to just consider it logically as follows:

You know how long it takes Leslie to complete her journey, so if you knew how far her journey was, you would have an answer immediately. Averagespeed = distance/time

You know how long it took Ann to complete her journey, and you know her average speed, so you can calculate the distance of her journey. Distance = Averagespeed*time

You know the two journeys were over the same distance as A to B distance must equal B to A distance. So now, knowing Ann's journey distance, you know Leslie's journey distance.

So finally, you have both distance and time for Leslie and can immediately calculate the average speed of her journey (as shown above).

edited Jul 25 at 23:28

answered Jul 25 at 23:27

Penguino

74449

A way to consider this type of problem while you are still grappling with the basics of algebra is to just consider it logically as follows:

You know how long it takes Leslie to complete her journey, so if you knew how far her journey was, you would have an answer immediately. Averagespeed = distance/time

You know how long it took Ann to complete her journey, and you know her average speed, so you can calculate the distance of her journey. Distance = Averagespeed*time

You know the two journeys were over the same distance as A to B distance must equal B to A distance. So now, knowing Ann's journey distance, you know Leslie's journey distance.

So finally, you have both distance and time for Leslie and can immediately calculate the average speed of her journey (as shown above).

edited Jul 25 at 23:28

answered Jul 25 at 23:27

Penguino

74449

edited Jul 25 at 23:28

answered Jul 25 at 23:27

Penguino

74449

answered Jul 25 at 23:27

Penguino

74449

answered Jul 25 at 23:27

Penguino

74449

add a commentÂ |Â

up vote
0
down vote

Leslie drive at speed $v_1.$ Anne drives at speed $v_2 = 48 frac mih$

Suppose $t$ hours pass from the time of departure to the time of meeting.

In that time Anne covers $v_2t$ miles and Leslie covers $v_1t$ miles.

over the rest of the journey each covers the distance that the other traveled and the times are given.

$frac v_1tv_2 = 4$ and $frac v_2tv_1 = 9$

Yow we can cross cross multiply each plug $v_2 = 48$ and sustitue from one equation to the other.

But I think would be more fun to divide one by the other.

$dfrac frac v_1tv_2frac v_2tv_1 = frac 49\
frac v_1^2v_2^2 = frac 49\
v_1 = frac 23 v_2$

answered Jul 25 at 23:34

Doug M

39k31749

add a commentÂ |Â

up vote
0
down vote

Leslie drive at speed $v_1.$ Anne drives at speed $v_2 = 48 frac mih$

Suppose $t$ hours pass from the time of departure to the time of meeting.

In that time Anne covers $v_2t$ miles and Leslie covers $v_1t$ miles.

over the rest of the journey each covers the distance that the other traveled and the times are given.

$frac v_1tv_2 = 4$ and $frac v_2tv_1 = 9$

Yow we can cross cross multiply each plug $v_2 = 48$ and sustitue from one equation to the other.

But I think would be more fun to divide one by the other.

$dfrac frac v_1tv_2frac v_2tv_1 = frac 49\
frac v_1^2v_2^2 = frac 49\
v_1 = frac 23 v_2$

answered Jul 25 at 23:34

Doug M

39k31749

add a commentÂ |Â

up vote
0
down vote

Leslie drive at speed $v_1.$ Anne drives at speed $v_2 = 48 frac mih$

Suppose $t$ hours pass from the time of departure to the time of meeting.

In that time Anne covers $v_2t$ miles and Leslie covers $v_1t$ miles.

over the rest of the journey each covers the distance that the other traveled and the times are given.

$frac v_1tv_2 = 4$ and $frac v_2tv_1 = 9$

Yow we can cross cross multiply each plug $v_2 = 48$ and sustitue from one equation to the other.

But I think would be more fun to divide one by the other.

$dfrac frac v_1tv_2frac v_2tv_1 = frac 49\
frac v_1^2v_2^2 = frac 49\
v_1 = frac 23 v_2$

answered Jul 25 at 23:34

Doug M

39k31749

Leslie drive at speed $v_1.$ Anne drives at speed $v_2 = 48 frac mih$

Suppose $t$ hours pass from the time of departure to the time of meeting.

In that time Anne covers $v_2t$ miles and Leslie covers $v_1t$ miles.

over the rest of the journey each covers the distance that the other traveled and the times are given.

$frac v_1tv_2 = 4$ and $frac v_2tv_1 = 9$

Yow we can cross cross multiply each plug $v_2 = 48$ and sustitue from one equation to the other.

But I think would be more fun to divide one by the other.

$dfrac frac v_1tv_2frac v_2tv_1 = frac 49\
frac v_1^2v_2^2 = frac 49\
v_1 = frac 23 v_2$

answered Jul 25 at 23:34

Doug M

39k31749

answered Jul 25 at 23:34

Doug M

39k31749

answered Jul 25 at 23:34

Doug M

39k31749

answered Jul 25 at 23:34

Doug M

39k31749

add a commentÂ |Â

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