1. ## Average speed

look at part c of the question. i accidently put that in. cause on the way there the speed is determined by the time so u just add both speeds and divide by two. but on the way back its determined by distance so u have to do a bunch of other math. but for part c.. how can the avg of the whole speed be72.5? i accidently put that in cause i thought the way there was 72.5 and back was 72.5 so that avg was obviously 72.5.. but since the way back was based on distance i had to rethink it... anyways how is the avg of the whole trip 72.5??

2. Hello, Legendsn3verdie!

We need this formulas: . $\text{Distance} \:=\:\text{Speed} \times \text{Time} \quad\Rightarrow\quad T \:=\:\frac{D}{S}$

Also: . $\text{Average speed} \:=\:\frac{\text{Total distance}}{\text{Total time}}$

I'll do the first two parts . . .

You drive from $A$ to $B$, half the time at 59 km/hr, and half the time at 86 km/hr.

On the way back, you drive half the distance at 59 km/hr, and half the distance at 86 km\hr.

(a) What is your average speed from $A$ to $B$?

Let $D$ = distance from $A$ to $B.$
Let $S$ = average speed. . Let $T$ = total time driving.
. . Then: . $D \:=\:S\cdot T\;\;{\color{blue}[1]}$

We drove at 59 km/hr for $\frac{1}{2}T$ hours.
. . The distance is: . $\frac{59}{2}T$ km.
We drove at 86 km/hr for $\frac{1}{2}T$ hours.
. . The distance is: . $\frac{86}{2}T$ km.
The total distance is: . $\frac{59}{2}T + \frac{86}{2}T \:=\:\frac{145}{2}T\text{ km.}\;\;{\color{blue}[2]}$

Equate [1] and [2]: . $S\!\cdot\!T \;=\;\frac{145}{2}T\quad\Rightarrow\quad S \:=\:72.5$

Therefore, the average speed from $A$ to $B$ is: . $\boxed{72.5\text{ km/hr}}$

You were right . . . You can average the two speeds.

(b) What is your average speed from $B$ to $A$?
This part is much trickier!

Let $D$ = distance from B to A.

We drove $\frac{1}{2}D$ km at 59 km/hr.
. . This took: . $\frac{\frac{1}{2}D}{59} \:=\:\frac{D}{118}$ hours.

We drove $\frac{1}{2}D$ km at 86 km/hr.
. . This took: . $\frac{\frac{1}{2}D}{86} \:=\:\frac{D}{172}$ hours.

We drove a total of: . $\frac{D}{118} + \frac{D}{172} \:=\:\frac{18D}{1247}$ hours.

We drove $D$ km in $\frac{18D}{1247}$ hours.

Our average speed is: . $\frac{D}{\frac{18D}{1247}} \;=\;\frac{1247}{18} \;\approx\;\boxed{69.27}\text{ km/hr}$

This time we can not average the two speeds!
.

3. For C) I got 71.221 as the average speed.

You drive from San Antonio to Houston half the time at 59km/h and the other half at 86 km/h.

So $59 \cdot \dfrac{t}{2} = \dfrac{59t}{2}$ km
and $86 \cdot \dfrac{t}{2} = {43t}$ km

We have so far the total distance: $\dfrac{145t}{2}$ km

Now on the way back, half the distance at 59km/h and the other at 86km/h.

So half of $\dfrac{145t}{2}$ km is $\dfrac{145t}{4}$ km. Hence $\dfrac{\frac{145t}{4}}{59}={\dfrac{145t}{236}}$ h is the time half way.

And $\dfrac{\frac{145t}{4}}{86}={\dfrac{145t}{344}}$ h is the time for the other half.

So our total distance traveled is:
$\left(\dfrac{59t}{2} + 43t + \dfrac{145t}{4} + \dfrac{145t}{4}\right)$ km

and our total time is:
$\left(\dfrac{t}{2} + \dfrac{t}{2} + \dfrac{145t}{236} + \dfrac{145t}{344}\right)$ h

So the avg. speed for entire trip would be:
$Avg. Speed={\dfrac{\frac{59t}{2} + 43t + \frac{145t}{4} + \frac{145t}{4}}{\frac{t}{2} + \frac{t}{2} + \frac{145t}{236} + \frac{145t}{344}}}={\dfrac{2942920}{41321}}\approx {71.221}$ km/h.

4. Originally Posted by Soroban
Hello, Legendsn3verdie!

We need this formulas: . $\text{Distance} \:=\:\text{Speed} \times \text{Time} \quad\Rightarrow\quad T \:=\:\frac{D}{S}$

Also: . $\text{Average speed} \:=\:\frac{\text{Total distance}}{\text{Total time}}$

I'll do the first two parts . . .

Let $D$ = distance from $A$ to $B.$
Let $S$ = average speed. . Let $T$ = total time driving.
. . Then: . $D \:=\:S\cdot T\;\;{\color{blue}[1]}$

We drove at 59 km/hr for $\frac{1}{2}T$ hours.
. . The distance is: . $\frac{59}{2}T$ km.
We drove at 86 km/hr for $\frac{1}{2}T$ hours.
. . The distance is: . $\frac{86}{2}T$ km.
The total distance is: . $\frac{59}{2}T + \frac{86}{2}T \:=\:\frac{145}{2}T\text{ km.}\;\;{\color{blue}[2]}$

Equate [1] and [2]: . $S\!\cdot\!T \;=\;\frac{145}{2}T\quad\Rightarrow\quad S \:=\:72.5$

Therefore, the average speed from $A$ to $B$ is: . $\boxed{72.5\text{ km/hr}}$

You were right . . . You can average the two speeds.

This part is much trickier!

Let $D$ = distance from B to A.

We drove $\frac{1}{2}D$ km at 59 km/hr.
. . This took: . $\frac{\frac{1}{2}D}{59} \:=\:\frac{D}{118}$ hours.

We drove $\frac{1}{2}D$ km at 86 km/hr.
. . This took: . $\frac{\frac{1}{2}D}{86} \:=\:\frac{D}{172}$ hours.

We drove a total of: . $\frac{D}{118} + \frac{D}{172} \:=\:\frac{18D}{1247}$ hours.

We drove $D$ km in $\frac{18D}{1247}$ hours.

Our average speed is: . $\frac{D}{\frac{18D}{1247}} \;=\;\frac{1247}{18} \;\approx\;\boxed{69.27}\text{ km/hr}$

This time we can not average the two speeds!
.
We drove $D$ km in $\frac{18D}{1247}$ hours.

how did u reduce that down to 18d/1247??

(d/118) + ( d/172) = 290d/20296 = time

speed = Distance/(290d/20296)

speed = 20296/290

speed= 69.986

5. Originally Posted by Pn0yS0ld13r
For C) I got 71.221 as the average speed.

You drive from San Antonio to Houston half the time at 59km/h and the other half at 86 km/h.

So $59 \cdot \dfrac{t}{2} = \dfrac{59t}{2}$ km
and $86 \cdot \dfrac{t}{2} = {43t}$ km

We have so far the total distance: $\dfrac{145t}{2}$ km

Now on the way back, half the distance at 59km/h and the other at 86km/h.

So half of $\dfrac{145t}{2}$ km is $\dfrac{145t}{4}$ km. Hence $\dfrac{\frac{145t}{4}}{59}={\dfrac{145t}{236}}$ h is the time half way.

And $\dfrac{\frac{145t}{4}}{86}={\dfrac{145t}{344}}$ h is the time for the other half.

So our total distance traveled is:
$\left(\dfrac{59t}{2} + 43t + \dfrac{145t}{4} + \dfrac{145t}{4}\right)$ km

and our total time is:
$\left(\dfrac{t}{2} + \dfrac{t}{2} + \dfrac{145t}{236} + \dfrac{145t}{344}\right)$ h

So the avg. speed for entire trip would be:
$Avg. Speed={\dfrac{\frac{59t}{2} + 43t + \frac{145t}{4} + \frac{145t}{4}}{\frac{t}{2} + \frac{t}{2} + \frac{145t}{236} + \frac{145t}{344}}}={\dfrac{2942920}{41321}}\approx {71.221}$ km/h.

hm whatever you did is hard for me to understand..

6. Just to weigh in here since we have two answers, as much as I hate to side against Soroban, Pn0yS0ld13r has it right. This is a "mind trap" question (which others might call a trick question.) The trap is that we cannot assume that the return trip takes the same amount of time as the trip out. If you assume the times are the same the you get the OP's answer.

-Dan