0
$\begingroup$

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?

$\endgroup$

5 Answers 5

Reset to default
1
$\begingroup$

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:

$\frac{4}{t} = \frac{t}{9}$

$t^2 = 36$

$t = 6$ hours

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

Leslie's speed is therefore $\frac{192}{6} = 32$ mph

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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).

$\endgroup$
0
$\begingroup$

Leslie drive at speed $v_1.$ Anne drives at speed $v_2 = 48 \frac {mi}{h}$

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.

$\frac {v_1t}{v_2} = 4$ and $\frac {v_2t}{v_1} = 9$

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.

$\dfrac {\frac {v_1t}{v_2}}{\frac {v_2t}{v_1}} = \frac {4}{9}\\ \frac {v_1^2}{v_2^2} = \frac 49\\ v_1 = \frac 23 v_2$

$\endgroup$

You must log in to answer this question.