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November 1st, 2019, 07:03 PM  #1 
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 178 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 $v_{b}=\textrm{blue car}$ $v_{r}=8$ $v_{b}=t5$ Since $v=\dfrac{dx}{dt}$ Then: $\dfrac{dx}{dt}=8$ $x(t)=8t+c$ $x(0)=0\,, c=0$ $x(t)_{r}=8t$ $\dfrac{dx}{dt}=t5$ $x(t)=\frac{t^2}{2}5t+c$ $x(0)=0\, c=0$ $x(t)_{b}= \frac{t^2}{2}5t$ So by equating both I could obtain the time isn't it?. $8t=\frac{t^2}{2}5t$ $0=\frac{t^2}{2}13t$ $0=t(t26)$ 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: 3,093 Thanks: 1675 
$8t = \dfrac{1}{2}(t5)^2$ $16t = t^210t+25$ $0 = t^226t+25$ $0=(t25)(t1)$ 
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 $v_{cyan} = t5$. So, $8 \cdot t_c = (t_c5) \cdot (t_c  5)/2$ Which means there exists a universe where one catches the other at the very first second. 
November 1st, 2019, 08:18 PM  #4  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus  Quote:
Wouldn't it be $\left(v8\right)$ ? Can you help me with that part?.  
November 1st, 2019, 08:22 PM  #5  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus  Quote:
What's the meaning of treating it as a step function?. How would that change your solution?.  
November 1st, 2019, 08:25 PM  #6  
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle  Quote:
$v(t)$ = velocity at time $t$. $s(t)$ = distance at time $t$. And you can ignore $constant$ part.  
November 1st, 2019, 08:39 PM  #7  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 178 Thanks: 5 Math Focus: Calculus  Quote:
$A=\frac{bh}{2}$ From this because the triangle is isoceles one can infer that the area is: $A=\frac{\left(t5\right)^2}{2}$ 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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