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October 21st, 2019, 08:39 PM  #1 
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  How do I find the instant when two moving objects are together using a graph?
The problem is as follows: The graph from below describes the motion of an electric car and a diesel truck. It is known that they pass through the same point at $t=0$. Find the instant when they are together again. The alternatives in my book are as follows: $\begin{array}{ll} 1.&\textrm{12 s}\\ 2.&\textrm{15 s}\\ 3.&\textrm{18 s}\\ 4.&\textrm{16 s}\\ 5.&\textrm{20 s} \end{array}$ For this problem the only thing that I could come up with was to identify the two equations for speed for both the truck and the electric car which are as follows. I'm using the labels $\textrm{t=truck and c=electric car}$ $v_{t}(t)=10$ $v_{c}(t)=2t$ from equating both I obtainted the time when they do have the same speed. (For brevity purposes I'm omitting the units but they are consistent) $2t=10$ $t=5$ But from then on I don't know what to do with that information or how do I join it with other equation to get to the time when they are together. Can somebody help me with this?. 
October 21st, 2019, 09:26 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 
Area under the velocity graph is displacement ... truck displacement = car displacement $\displaystyle \int_0^t 10 \, dt = \int_0^6 2t \, dt + \int_6^t 12 \, dt$ $10t = 36 + (12t72)$ $10t = 12t 36 \implies t=18$ 
October 22nd, 2019, 06:37 AM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 
w/o calculus for $0 \le t \le 6$ ... $v_c = 2t \implies x_c(6) = \dfrac{1}{2}(2) \cdot 6^2 = 36$ $v_T = 10 \implies x_T(6) = 10(6) = 60$ truck is 24 meters ahead at time $t=6$ for $t > 6$ $24 + 10t = 12t \implies t = 12$ car passes truck at time $t= 6 + 12$ sec. 
October 28th, 2019, 09:14 PM  #4  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Quote:
$\dfrac{1}{2}(2) \cdot 6^2 = 36$ Are you implying to use the known areas by the area of the right triangle but that wouldn't be? $A=\frac{bh}{2}$ I'd like you could explain me this part please.  
October 29th, 2019, 04:50 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 21,124 Thanks: 2332 
The sketch isn't to scale. In the first 6 s, the car's average velocity is 6 m/s, so the truck gets ahead by 6(10  6) m = 24 m. After 6 seconds, the car's velocity relative to the truck is a constant 2m/s, so the car catches up with the truck after a further 12 seconds. Last edited by skipjack; October 29th, 2019 at 05:14 AM. 
October 29th, 2019, 08:00 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 3,101 Thanks: 1677 
Velocity graphs to scale. Compare the area of the red trapezoid to the blue rectangle 
November 1st, 2019, 02:41 PM  #7  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Quote:
The term average is where I'm stuck at. The second statement I presume it comes from using $6t$. The third part about relative velocity of the car where you mention that is constant I'm lost. What's the meaning of relative velocity and how did you obtained?. Sorry but these look conceptual questions and I'd like some guidance because I think it is the answer of how to understand what you mentioned.  
November 1st, 2019, 02:56 PM  #8  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Quote:
The part where you used this equation: $10t+24=12t$ Is only possible when the two objects are moving with no acceleration involved, in other words uniform motion, hence this is why you used the interval $t>6$. But again, I think that for that equation from above, areas were not used. Am I right?.  
November 1st, 2019, 03:00 PM  #9 
Senior Member Joined: Jun 2019 From: USA Posts: 386 Thanks: 211 
The average value of any f(t) is $\displaystyle \bar{f} = \frac{1}{\Delta t} \int_{t_0}^{t_0 + \Delta t} f dt$. An integral is the area under the curve. In the first 6 seconds, the average velocity is the area under the triangle divided by the width of the triangle. This is the "average height." For a linear relation like this, it's easy. The average velocity will be $\displaystyle \frac{0~m/s+12~m/s}{2} = 6~m/s$. 
November 1st, 2019, 05:25 PM  #10  
Senior Member Joined: Jun 2017 From: Lima, Peru Posts: 188 Thanks: 5 Math Focus: Calculus  Quote:
 

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