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April 11th, 2014, 01:37 PM  #1 
Senior Member Joined: Jan 2012 Posts: 739 Thanks: 7  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. 
April 11th, 2014, 01:56 PM  #2  
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra  Quote:
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.  
April 11th, 2014, 03:49 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, Chikis! Quote:
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 $ $\displaystyle \;\;\;x^27x8 \;=\;0$ $\displaystyle \;\;\;(x8)(x+1) \;=\;0$ $\displaystyle \;\;\;\begin{Bmatrix}x8 \:=\:0 & \Rightarrow & x \:=\:8 \\ x+1 \:=\:0 & \Rightarrow & \rlap{///////}x \:=\:\text{}1 \end{Bmatrix}$ His average speed for the first 4 km was $\displaystyle \boxed{8\text{ km/hr.}}$  
April 15th, 2014, 02:07 PM  #4 
Senior Member Joined: Jan 2012 Posts: 739 Thanks: 7 
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?

April 16th, 2014, 02:27 AM  #5 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 
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?! 
February 28th, 2019, 04:01 PM  #6 
Newbie Joined: Feb 2019 From: Benin city, Edo state, Nigeria. Posts: 1 Thanks: 0  Average speed question.
Total distance=9km. Total time=1hr. Let speed be x For the first 4km, x=4/t, t=4/x. For 5km, x+2=5/t, t=5/x+2 hrs. Substituting t=5/x+2 in t=4/x , 5/x+2=4/x, cross multiply 5x=4x+8. X=8km/hr.

February 28th, 2019, 04:24 PM  #7 
Senior Member Joined: Aug 2012 Posts: 2,260 Thanks: 687  

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