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December 9th, 2017, 05:22 PM  #1 
Member Joined: Apr 2012 Posts: 72 Thanks: 3  How big is the speed of y from the speed of x? Last edited by mathLover; December 9th, 2017 at 05:48 PM. 
December 9th, 2017, 05:22 PM  #2 
Member Joined: Apr 2012 Posts: 72 Thanks: 3 
I sketched the timeline on paper, and I think the answer is 2, but I do not know how to prove it formally

December 10th, 2017, 04:56 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Better English would be "How much larger is the speed of y than the speed of x?" And I take that "the distance between x and y was four times smaller" means "four times smaller" (1/4) than the original distance between A and B. Let d be the distance between A and B, $\displaystyle v_x$ the speed of x and $\displaystyle v_y$ the speed of y. After a time t x will have gone distance $\displaystyle v_xt$ from A and y $\displaystyle v_yt$ from B. The distance between them is $\displaystyle d v_xt v_yt$. x and y will meet when $\displaystyle d v_xt_1 v_yt_1= 0$ or $\displaystyle (v_x+ v_y)t_1= d$, $\displaystyle t_1= \frac{d}{v_x+ v_y}$ so their distance from A is $\displaystyle \frac{dv_x}{v_x+ v_y}$. At time $\displaystyle t_2$ the distance between x and y is 1/4 the original distance between A and B, d/4. So $\displaystyle d v_xt_2 v_yt_2= d/4$ or $\displaystyle (v_x+ v_y)t_2= 3d/4$, [math]t_2= \frac{3d}{4(v_x+ v_y)}. At that time, z starts running from A to B with speed $\displaystyle v_z$. z must meet x and y at the same point at which x and y meet so at the same time: $\displaystyle t_1$. When the three meet, z will have run at speed $\displaystyle v_z$ for time $\displaystyle t_1 t_2= \frac{d}{4(v_x+ v_y)}$ so will have gone distance $\displaystyle \frac{v_zd}{4(v_x+ v_y)}$. Since all three meet at the same point, with distance from A $\displaystyle \frac{v_zd}{4(v_x+ v_y)}$. Before we had that the distance from A to the point where x and y meet was $\displaystyle \frac{dv_x}{v_x+ v_y}$ so we must have $\displaystyle \frac{dv_x}{v_x+ v_y}= \frac{dv_z}{4(v_x+ v_y)}$. We can immediately cancel the "d" terms and the "$\displaystyle v_x+ v_y$" terms leaving $\displaystyle v_x= \frac{v_z}{4}$. Finally y arrives at A having run distance d, so at time $\displaystyle t_3= \frac{d}{v_y}$ at the same time z will have run from A to B, also distance d, at the same time, $\displaystyle t_3$ so having run for time $\displaystyle t_3 t_2: t_3 t_2= \frac{d}{v_z}$. Putting those together, $\displaystyle t_3= t_2+ \frac{d}{v_z}= \frac{d}{v_y}$. $\displaystyle \frac{d}{v_z}= \frac{d}{v_y} t_2$. $\displaystyle \frac{d}{v_y}= \frac{d}{v_z}+ t_2$. $\displaystyle d= v_y(\frac{d}{v_z}+ t_2)$. $\displaystyle v_y= \frac{d}{\frac{d}{v_z}+ t_2}$. So we now have $\displaystyle v_x= \frac{v_z}{4}$ and $\displaystyle v_y= \frac{d}{\frac{d}{v_z}+ t_2}$. Above, we had $\displaystyle t_2= \frac{3d}{4(v_x+ v_y)}$. Replacing $\displaystyle v_x$ and $\displaystyle v_y$ with the formulas above, we have $\displaystyle t_2= \frac{3d}{4(\frac{v_z}{4}+ \frac{d}{\frac{d}{v_z}+ t_2})}$. Here the algebra gets a little complicated! This works out to a quadratic equation for $\displaystyle t_2$ which I will leave it to you to solve. Once you have both $\displaystyle v_y$ and $\displaystyle v_x$ as multiples of $\displaystyle v_z$ the ratio will be easy. If we can get $\displaystyle t_2$ in terms of $\displaystyle v_z$ only we can find $\displaystyle \frac{v_y}{v_x}$. Last edited by skipjack; December 10th, 2017 at 05:04 AM. 
December 10th, 2017, 09:12 AM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,115 Thanks: 913 
YIKES CB, I've never seen such a solution for a distance = speed * time problem ! I'm overly impressed... Do you agree that the problem could be worded this way: We have straight path X to Y, distance = 4d miles. A leaves from X at same time as B leaves from Y. When A and B are d miles apart, C leaves from X. Later, A, B and C all meet at the same point. Later, B reaches X at same time as C reaches Y. What is B's speed in terms of A's speed? Solving: Let a mph = A's speed, b mph = B's speed, c mph = C's speed. What sayest thou? 
December 10th, 2017, 10:19 AM  #5 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,115 Thanks: 913 
The "meeting time" is simply 4d / (a + b); agree?

December 11th, 2017, 05:53 AM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,115 Thanks: 913 
Here's an integer example: 4d = 300 (distance X to Y) a = 5, b = 10, c = 20 u = 15 (time when A and B are d apart) v = 20 (time when meeting occurs) w = 30 (time when B reaches X and C reaches Y) 
December 11th, 2017, 04:01 PM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,115 Thanks: 913 
Answer: b = 2a (got some help! The "story": ([time when B reaches X]  [time when A and B are d apart])*c = 4d (4d/b  3d/(a+b))c = 4d 4/b  3/(a+b) = 4/c 4ac + bc = 4ab + 4b^2 [eq. 1] [A's distance until meeting] = [C's distance until meeting] a(4d/(a+b)) = c(4d/(a+b)  3d/(a+b)), 4a/(a+b) = c/(a+b) c = 4a [eq. 2] Substitute in [eq. 1] to get: 16a^2 + 4ab = 4ab + 4b^2. 4a^2 = b^2 b = 2a 

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