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 November 1st, 2019, 07:03 PM #1 Senior Member     Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus Is it possible to find when two cars will meet given this graph? The problem is as follows: The figure from below shows the speed against time of two cars, one blue and the other orange. It is known that both depart from the same spot. Find the instant on seconds when one catches the other. The given alternatives are: $\begin{array}{ll} 1.&14\,s\\ 2.&16\,s\\ 3.&20\,s\\ 4.&25\,s\\ 5.&28\,s\\ \end{array}$ For this problem I attempted to do the "trick" using the areas behind the curves but I couldn't find the answer. So far I could only state the equations as this: I'm using $v_{r}=\textrm{orange car}$ and $v_{b}=\textrm{blue car}$ $v_{r}=8$ $v_{b}=t-5$ Since $v=\dfrac{dx}{dt}$ Then: $\dfrac{dx}{dt}=8$ $x(t)=8t+c$ $x(0)=0\,, c=0$ $x(t)_{r}=8t$ $\dfrac{dx}{dt}=t-5$ $x(t)=\frac{t^2}{2}-5t+c$ $x(0)=0\, c=0$ $x(t)_{b}= \frac{t^2}{2}-5t$ So by equating both I could obtain the time isn't it?. $8t=\frac{t^2}{2}-5t$ $0=\frac{t^2}{2}-13t$ $0=t(t-26)$ So time would be $26\,s$. Supposedly the answer is $25$ and i'm off by one digit. But it doesn't seem to be in any of the alternatives given. Could it be that I'm not getting the right picture of this problem correctly?. Can somebody give me a help?.
 November 1st, 2019, 07:34 PM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 $8t = \dfrac{1}{2}(t-5)^2$ $16t = t^2-10t+25$ $0 = t^2-26t+25$ $0=(t-25)(t-1)$
 November 1st, 2019, 08:02 PM #3 Senior Member     Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle It is a matter of integration. Let's assume when $t=t_c$ they "catch", which means the distance they traveled is equal. We know $v_{orange}=8$ and $v_{cyan} = t-5$. So, $8 \cdot t_c = (t_c-5) \cdot (t_c - 5)/2$ Which means there exists a universe where one catches the other at the very first second.
November 1st, 2019, 08:18 PM   #4
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Math Focus: Calculus

Quote:
 Originally Posted by tahirimanov19 It is a matter of integration. Let's assume when $t=t_c$ they "catch", which means the distance they traveled is equal. We know $v_{orange}=8$ and $v_{cyan} = t-5$. So, $8 \cdot t_c = (t_c-5) \cdot (t_c - 5)/2$ Which means there exists a universe where one catches the other at the very first second.
I'm a bit stuck at why the height of the triangle in this case is also $(t_{c}-5)$?. Initially I thought it was not possible. I mean it is obvious the base is $(t_{c}-5)$ but how about the height.

Wouldn't it be $\left(v-8\right)$ ? Can you help me with that part?.

November 1st, 2019, 08:22 PM   #5
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Quote:
 Originally Posted by skeeter $8t = \dfrac{1}{2}(t-5)^2$ $16t = t^2-10t+25$ $0 = t^2-26t+25$ $0=(t-25)(t-1)$
Out of curiosity. I've been mentioned that this can be treated as an step function. Is this part of your solution?.

What's the meaning of treating it as a step function?. How would that change your solution?.

November 1st, 2019, 08:25 PM   #6
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Math Focus: Area of Circle
Quote:
 Originally Posted by Chemist116 I'm a bit stuck at why the height of the triangle in this case is also $(t_{c}-5)$?. Initially I thought it was not possible. I mean it is obvious the base is $(t_{c}-5)$ but how about the height. Wouldn't it be $(v-$?. Can you help me with that part?.
From graph $v_{cyan}(t) = t-5$, so $s_{cyan}(t) = \int v_{cyan}(t) dt = \int (t-5)dt=\dfrac{(t-5)^2}{2} + constant$.

$v(t)$ = velocity at time $t$.
$s(t)$ = distance at time $t$.
And you can ignore $constant$ part.

November 1st, 2019, 08:39 PM   #7
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Quote:
 Originally Posted by tahirimanov19 From graph $v_{cyan}(t) = t-5$, so $s_{cyan}(t) = \int v_{cyan}(t) dt = \int (t-5)dt=\dfrac{(t-5)^2}{2} + constant$. $v(t)$ = velocity at time $t$. $s(t)$ = distance at time $t$. And you can ignore $constant$ part.
I thought that you obtained the area not from an integration but from the area of the triangle.

$A=\frac{bh}{2}$

From this because the triangle is isoceles one can infer that the area is:

$A=\frac{\left(t-5\right)^2}{2}$

This was what I intended to ask in my previous question.

At this point the message I'm getting is subtract from the time $t$ the time from which the second object is departing. But should I say this is a norm?

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