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April 11th, 2014, 02:37 PM   #1
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His average speed for the first 4km is what?

A man runs a distance of 9km at a constant speed for the first 4km and then 2km/h faster for the rest of the distance. The whole run takes him one hour. His average speed for the first 4km is what?



Total distance = 9km
Total time = 1 h

For the 1st for 4km, he moved at a constant speed.

For the last 5km, his speed was increased by 2km/h.

Let the costant speed at 4km be xkm/h.

At the last 5km, his speed became (x + 2) km/h.


The total speed for the whole runs took him
[(x+2) + x] km/h
= (2x + 2) km/h

speed s = distance d/time t

s = d/t

t = 9/(2x + 2)


1 = 9/(2x + 2)

2x + 2 = 9

2x = 7

x = 7/2 = 3.5 km/h

Therefore the speed for the 4 km = 3.5 km/h

Since the speed was increased by 2km/h, that means the speed for the last 5 km/h would become (3.5 + 2)
= 5.5 km/h.

But when I checked my answer, my did not seem right. Please I need help. Thank you.
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April 11th, 2014, 02:56 PM   #2
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Quote:
Originally Posted by Chikis View Post
The total speed for the whole runs took him
[(x+2) + x] km/h
= (2x + 2) km/h
That’s not the way to calculate speed. Speeds are not additive.

Instead you should find the time taken to run the first $\mathrm{4\ km}$ at $x\ \mathrm{km\,h^{-1}}$ and the time the time taken to run the last $\mathrm{5\ km}$ at $(x+2)\ \mathrm{km\,h^{-1}}$, and add the two times.
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April 11th, 2014, 04:49 PM   #3
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Hello, Chikis!

Quote:
A man runs a distance of 9 km at a constant speed for the first 4 km
and then 2 km/hr faster for the rest of the distance.
The whole run takes him one hour.
What is his average speed for the first 4km?

We will use:$\displaystyle \;\;\text{Time} \;=\;\frac{\text{Distance}}{\text{Speed}}$

Let $\displaystyle x$ = average speed for the first 4 km.

He ran 4 km at $\displaystyle x$ km/hr.
This took:$\displaystyle \;\frac{4}{x}$ hours.

Then he ran 5 km at $\displaystyle (x+2)$ km/hr.
This took:$\displaystyle \;\frac{5}{x+2}$ hours.

His total time is one hour.
$\displaystyle \;\;\;\frac{4}{x} + \frac{5}{x+2} \;=\;1$


Multiply by $\displaystyle x(x+2)$:

$\displaystyle \;\;\;4(x+2) + 5x \;=\;x(x+2)$

$\displaystyle \;\;\;4x + 8 + 5x \;=\;x^2+2x $

$\displaystyle \;\;\;x^2-7x-8 \;=\;0$

$\displaystyle \;\;\;(x-8)(x+1) \;=\;0$

$\displaystyle \;\;\;\begin{Bmatrix}x-8 \:=\:0 & \Rightarrow & x \:=\:8 \\
x+1 \:=\:0 & \Rightarrow & \rlap{///////}x \:=\:\text{-}1 \end{Bmatrix}$

His average speed for the first 4 km was $\displaystyle \boxed{8\text{ km/hr.}}$

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April 15th, 2014, 03:07 PM   #4
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My observation here is that we can only make equation for the total time and use it to find the speed. Why is it that if we make equation for the total distance (as I did in my previous work), we cannot use it to get the accurate value for x the speed?
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April 16th, 2014, 03:27 AM   #5
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As I pointed out, your mistake was adding up speeds. You don’t that – it’s meaningless. It’s like saying if you travel one part of your journey at 40 miles per hour and the other part at 30 miles per hour the “total speed” for your journey is 70 miles per hour?!
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