Average Speed

[HelenA]

Newport and Vernonville are 192 miles apart. A car leaves Newport travelling towards Vernonville, and another car leaves Vernonville travelling to Newport at the same time. The car leaving Newport averages 10 miles per hour more than the hour, and both car meet up after 1 hour and 36 minutes. What are the average speeds of the cars?

** My mind has gone blank on the correct application of D= r * t. ** Help is greatly appreciated as it has been almost 18 + years since I have done this application of Algebra. Thanks.

 

[soroban]

Hello, HelenA!

Newport and Vernonville are 192 miles apart.
A car leaves N travelling towards V, and another car leaves V towards Newport at the same time.
The car leaving N averages 10 mph more than the other, and both cars meet after 1 hour, 36 minutes.
What are the average speeds of the cars?

\(\displaystyle \text{Let: }\;\begin{Bmatrix}x &=& \text{speed of the car from }V\\ x+10 &=& \text{speed of the car from }N \end{Bmatrix}\)

\(\displaystyle \text{Their combined speed is: }\:x+(x+10)\text{ mph.}\)
. . \(\displaystyle \text{That is, they are approaching each other at }2x+10\text{ mph.}\)

\(\displaystyle \text{In 1 hour, 36 minutes }\left(\tfrac{8}{5}\text{ hours}\right),\,\text{ they covered 192 miles.}\)

\(\displaystyle \text{There is our equation: }\;\overbrace{192}^D \;=\;\overbrace{(2x+10)}^r\cdot\overbrace{\left(\tfrac{8}{5}\right)}^t\)

. . Got it?

 

[HelenA]

Just one question - how did you convert 1 hr and 36 minutes into 8/5? I understood everything else except that part. Thanks so much, again, for your help.

 

[BigGlenntheHeavy]

\(\displaystyle Another \ way: \ We \ know \ (or \ should) \ that \ R \ X \ T \ = \ D.\)

\(\displaystyle Hence, \ \ \ \ R \ \ \ \ \ \ \ X \ \ \ \ T \ \ \ = \ \ \ D\)

\(\displaystyle Car \ A: \ (y+10) \ X \ (1.6) \ \ = \ \ \ x\)

\(\displaystyle Car \ B: \ \ \ y \ \ \ \ \ \ \ X \ \ (1.6) \ \ = \ 192-x\)

\(\displaystyle Adding \ A \ and \ B, \ we \ get \ -> \ 32y+16 \ = \ 192, \ \implies \ y \ = \ 55\)

\(\displaystyle Hence, \ A \ = \ 65 \ mph \ and \ B \ = \ 55 \ mph\)

\(\displaystyle I'll \ leave \ the \ check \ up \ to \ whoever.\)

 

[Deleted member 4993]

Just one question - how did you convert 1 hr and 36 minutes into 8/5? I understood everything else except that part. Thanks so much, again, for your help.

1 hr and 36 minutes = 1 hr + (36/60) hours = 1 hr + (3/5) hr = 8/5 hr