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Two trains started moving one from point A and the other from point B against each other at the same time for a trip length of 240 km. The first train speed is 80 km/h and the second one is 100 km/h. What is the distance between the two trains 30 minutes before they meet?

 Jan 9, 2020
 #1
avatar+11331 
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80km/h*0.5h+100km/h*0.5h=40km+50km=90km     (30min=0.5h)

laugh

 Jan 9, 2020
edited by Omi67  Jan 9, 2020
 #2
avatar+24089 
+2

Two trains started moving one from point A and the other from point B against each other at the same time
for a trip length of 240 km.
The first train speed is 80 km/h and the second one is 100 km/h.
What is the distance between the two trains 30 minutes before they meet?

 

\(\begin{array}{|rcll|} \hline y_B &=& -100t+240 \\ y_A &=& 80t \\ \hline \end{array}\)

 

meeting time

\(\begin{array}{|rcll|} \hline \mathbf{y_A} &=& \mathbf{y_B} \\\\ 80t_{meet} &=& -100t_{meet}+240 \\ 180t_{meet} &=& 240 \\ t_{meet} &=& \dfrac{240}{180} \\ \mathbf{t_{meet}} &=& \mathbf{ \dfrac{4}{3}\ h } \qquad ( t_{meet}= 80\ \text{minutes} )\\ \hline \end{array}\)

 

\(t_{meet}- 30\ \text{minutes} = 50\ \text{minutes}\)

\(\begin{array}{|rcll|} \hline y_B &=& -100 \left(\dfrac{50}{60}\right)+240 \\ y_A &=& 80\left(\dfrac{50}{60}\right) \\ \hline y_B - y_A &=& -100 \left(\dfrac{5}{6}\right)+240 - 80\left(\dfrac{5}{6}\right) \\ &=& -180 \left(\dfrac{5}{6}\right)+240 \\ &=& -150+240 \\ &=& \mathbf{90\ \text{km}} \\ \hline \end{array}\)

 

The distance between the two trains 30 minutes before they meet is \(\mathbf{90\ \text{km}}\)

 

laugh

 Jan 9, 2020
 #3
avatar+20217 
+2

I would just say that their closing speed is 100 + 80 = 180 km/hr

    30 minutes (which is 1/2 hour)  before they meet would be 1/2 of this  = 90 km

 Jan 9, 2020

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