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3 years ago

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Lewiston and Vernonville are 208 miles apart. A car leaves Lewiston traveling towards Vernonville, and another car leaves Verno

nville 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?

1 answer:

iragen [17]3 years ago

3 0

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 $v_{A} =\frac{x_{A} }{t_{A} }$ (Equation A) and Average speed of the car B is $v_{B} =\frac{x_{B} }{t_{B} }$ (Equation B), where $x_{A}$ and $x_{B}$ are the distances and $t_{A}$ and $t_{B}$ 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

$96 minutes . \frac{1 hour}{60 minutes} = 1.6 h$

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

$v_{A} = \frac{208-x}{1.6h} + 10\frac{miles}{h}$ (Equation C)

$v_{B} = \frac{208-x}{1.6h}$ (Equation D)

From Equation A and C, we have:

$\frac{208-x}{1.6}+10 = \frac{x}{1.6}$

208-x+16 = x

208 + 16 = 2x

$x = \frac{224}{2}$

x = 112 miles

Replacing x in Equation A:

$v_{A} = \frac{112miles}{1.6h}$

$v_{A} = 70 miles per hour$

Replacing x in Equation B:

$v_{B} = \frac{208miles-112miles}{1.6h}$

$v_{B} = \frac{96miles}{1.6h}$

$v_{B} = 60 miles per hour$

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