1. ## Average Speed

A car travels from Mayville to Rome at an average speed of 30 miles per hour and returns immediately along the same route at an average speed of 40 miles per hour. Of the following, which is the closest to the average speed, in MPH, for the round-trip?

A. 32.0
B. 33.0
C. 34.3
D. 35.5
E. 36.5

(30 + 40)/2 = 35

So, 35 is the average speed for the round-trip.
From the choices given, the closest to 35 is 35.5.

2. ## Re: Average Speed

Originally Posted by harpazo
A car travels from Mayville to Rome at an average speed of 30 miles per hour and returns immediately along the same route at an average speed of 40 miles per hour. Of the following, which is the closest to the average speed, in MPH, for the round-trip?

A. 32.0
B. 33.0
C. 34.3
D. 35.5
E. 36.5

(30 + 40)/2 = 35

So, 35 is the average speed for the round-trip.
From the choices given, the closest to 35 is 35.5.
The idea is this: $r\cdot t=D$ The distance $D$ is constant.
Going $30\cdot t_1=D$ and coming back we have $40\cdot t_2=D$
Thus $t_1=\dfrac{D}{30}~\&~t_2=\dfrac{D}{40}$
You want $Av\cdot(t_1+t_2)=2D$
Can you solve for $Av~?$

3. ## Re: Average Speed

Originally Posted by harpazo
A car travels from Mayville to Rome at an average speed of 30 miles per hour and returns immediately along the same route at an average speed of 40 miles per hour. Of the following, which is the closest to the average speed, in MPH, for the round-trip?

A. 32.0
B. 33.0
C. 34.3
D. 35.5
E. 36.5

(30 + 40)/2 = 35

So, 35 is the average speed for the round-trip.
From the choices given, the closest to 35 is 35.5.
Your approach assumes that the car traveled at the two different speeds for the same amount of time, but it is the distance traveled at the two different speeds that is the same, not the time traveled. I would use a weighted average here:

$\displaystyle \frac{30t_1+40t_2}{t_1+t_2}=\overline{v}$

Note: this is just average speed being total distance over total time.

Let $\displaystyle D$ be the one way distance and we may write:

$\displaystyle D=30t_1=40t_2\implies t_1=\frac{4}{3}t_2$

Hence:

$\displaystyle \overline{v}=\frac{80t_2}{\dfrac{7}{3}t_2}=?$

4. ## Re: Average Speed

$d$ is one way distance

$v_1$ is initial trip speed

$v_2$ is return trip speed

$\bar{v} = \dfrac{\text{total distance}}{\text{total time}} = \dfrac{d+d}{\frac{d}{v_1}+\frac{d}{v_2}} = \dfrac{2d}{\frac{v_1d+v_2d}{v_1 \cdot v_2}} = \dfrac{2d(v_1 \cdot v_2)}{(v_1+v_2)d} = \dfrac{2(v_1 \cdot v_2)}{v_1+v_2}$

5. ## Re: Average Speed

Originally Posted by harpazo
A car travels from Mayville to Rome at an average speed of 30 miles per hour and returns immediately along the same route at an average speed of 40 miles per hour. Of the following, which is the closest to the average speed, in MPH, for the round-trip?

A. 32.0
B. 33.0
C. 34.3
D. 35.5
E. 36.5

(30 + 40)/2 = 35

So, 35 is the average speed for the round-trip.
From the choices given, the closest to 35 is 35.5.
Imagine this scenario:
You are travelling a distance of 1000 miles. You travel the first 999 miles by car at 100mph. You then walk the last mile (at 1mph).
Is your average speed (100+1)/2 = 50.5mph.
NO! Because you drove for a lot longer than you walked. Your average speed would be a lot closer to 100mph than to 10mph.

In your problem, you travel at 30mph for longer than you do at 40mph (going the same distance at a slower speed will take longer), so your answer will be closer to 30 than to 40.
(ie answers D and E are out!)

So, what you haven't done is to take time into account.

Facts: $\displaystyle Speed =\frac{distance}{time}$ which means that $\displaystyle Time = \frac{distance}{speed}$
and
$\displaystyle Average speed =\frac{total distance}{total time}$

To set up the equation let the distance there and the distance back both be d, then total distance is 2d.

Time at 30mph = $\displaystyle \frac{d}{30}$ while time at 40mph =$\displaystyle \frac{d}{40}$.

So total time is $\displaystyle \frac{d}{30}+\frac{d}{40} =\frac{40d+30d}{1200}=\frac{70d}{1200} =\frac{7d}{120}$

So, Average speed = $\displaystyle \frac{total distance}{total time} = \frac{2d}{\frac{7d}{120}} = 2d * \frac{120}{7d}=\frac{240}{7}\approx 34.3$

(notice how the d's cancelled out)

6. ## Re: Average Speed

Originally Posted by Plato
The idea is this: $r\cdot t=D$ The distance $D$ is constant.
Going $30\cdot t_1=D$ and coming back we have $40\cdot t_2=D$
Thus $t_1=\dfrac{D}{30}~\&~t_2=\dfrac{D}{40}$
You want $Av\cdot(t_1+t_2)=2D$
Can you solve for $Av~?$
AV = 2D/(t_1 + t_2)

7. ## Re: Average Speed

Originally Posted by MarkFL
Your approach assumes that the car traveled at the two different speeds for the same amount of time, but it is the distance traveled at the two different speeds that is the same, not the time traveled. I would use a weighted average here:

$\displaystyle \frac{30t_1+40t_2}{t_1+t_2}=\overline{v}$

Note: this is just average speed being total distance over total time.

Let $\displaystyle D$ be the one way distance and we may write:

$\displaystyle D=30t_1=40t_2\implies t_1=\frac{4}{3}t_2$

Hence:

$\displaystyle \overline{v}=\frac{80t_2}{\dfrac{7}{3}t_2}=?$
80 ÷ 7/3 = 34.2857142857, which is then rounded to two decimal places to become 34.3 = choice C.

8. ## Re: Average Speed

Originally Posted by Debsta
Imagine this scenario:
You are travelling a distance of 1000 miles. You travel the first 999 miles by car at 100mph. You then walk the last mile (at 1mph).
Is your average speed (100+1)/2 = 50.5mph.
NO! Because you drove for a lot longer than you walked. Your average speed would be a lot closer to 100mph than to 10mph.

In your problem, you travel at 30mph for longer than you do at 40mph (going the same distance at a slower speed will take longer), so your answer will be closer to 30 than to 40.
(ie answers D and E are out!)

So, what you haven't done is to take time into account.

Facts: $\displaystyle Speed =\frac{distance}{time}$ which means that $\displaystyle Time = \frac{distance}{speed}$
and
$\displaystyle Average speed =\frac{total distance}{total time}$

To set up the equation let the distance there and the distance back both be d, then total distance is 2d.

Time at 30mph = $\displaystyle \frac{d}{30}$ while time at 40mph =$\displaystyle \frac{d}{40}$.

So total time is $\displaystyle \frac{d}{30}+\frac{d}{40} =\frac{40d+30d}{1200}=\frac{70d}{1200} =\frac{7d}{120}$

So, Average speed = $\displaystyle \frac{total distance}{total time} = \frac{2d}{\frac{7d}{120}} = 2d * \frac{120}{7d}=\frac{240}{7}\approx 34.3$

(notice how the d's cancelled out)
Thank you for breaking it down piece by piece.