Lewiston and Vernonville are 208 miles apart. A car leaves Lewiston traveling towards​ Vernonville, and another car leaves Vernonville at the same​ time, traveling towards Lewiston. The car leaving Lewiston averages 10 miles per hour more than the​ other, and they meet after 1 hour and 36 minutes. What are the average speeds of the​ cars?

Respuesta :

Answer:

Average speed of the car A = 70 miles per hour

Average speed of the car B = 60 miles per hour

Explanation:

Average speed of the car A is [tex]v_{A} =\frac{x_{A} }{t_{A} }[/tex] (Equation A) and Average speed of the car B is [tex]v_{B} =\frac{x_{B} }{t_{B} }[/tex] (Equation B), where [tex]x_{A}[/tex] and [tex]x_{B}[/tex] are the distances and [tex]t_{A}[/tex] and [tex]t_{B}[/tex] are the times at which are travelling the cars A and B respectively.

We have to convert the time to the correct units:

1 hour and 36 minutes = 96 minutes

[tex]96 minutes . \frac{1 hour}{60 minutes} = 1.6 h[/tex]

From the diagram (Please see the attachment), we can see that at the time they meet, we have:

[tex]v_{A} = \frac{208-x}{1.6h} + 10\frac{miles}{h}[/tex] (Equation C)

[tex]v_{B} = \frac{208-x}{1.6h}[/tex] (Equation D)

From Equation A and C, we have:

[tex]\frac{208-x}{1.6}+10 = \frac{x}{1.6}[/tex]

208-x+16 = x

208 + 16 = 2x

[tex]x = \frac{224}{2}[/tex]

x = 112 miles

Replacing x in Equation A:

[tex]v_{A}  = \frac{112miles}{1.6h}[/tex]

[tex]v_{A} = 70 miles per hour[/tex]

Replacing x in Equation B:

[tex]v_{B}  = \frac{208miles-112miles}{1.6h}[/tex]

[tex]v_{B}  = \frac{96miles}{1.6h}[/tex]

[tex]v_{B}  = 60 miles per hour[/tex]

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