Question

Two cars simultaneously left Points A and B and headed towards each other, and met after 2 hours and 45 minutes. The distance between points A and B is 264 miles. What is the speeds of the cars, if one of the cars travels 14 mph faster than the other?
Please help
I've done all my homework and this is the only problem I can't understand

Answer 1

Hello!

The answers are:

$FirstCarSpeed=41mph\\SecondCarSpeed=55mph$

Why?

To calculate the speed of the cars, we need to write two equations, one for each car, in order to create a relation between the two speeds and be able to calculate one in function of the other.

So,

Tet be the first car speed "x" and the second car speed "y", writing the equations we have:

For the first car:

$x_{FirstCar}=x_o+v*t$

For the second car:

We know that the speed of the second car is the speed of the first car plus 14 mph, so:

$x_{SecondCar}=x_o+(v+14mph)*t$

Now, from the statement that both cars met after 2 hours and 45 minutes, and the distance between to cover (between A and B) is 264 miles,  so, we can calculate the relative speed between them:

If the cars are moving towards each other the relative speed will be:

$RelativeSpeed=FirstCarSpeed-(-SecondCarspeed)\\\\RelativeSpeed=x-(-x-14mph)=2x+14mph$

Then, since we know that they covered a combined distance equal to 264 miles in 2 hours + 45 minutes, we have:

$2hours+45minutes=120minutes+45minutes=165minutes\\\\\frac{165minutes*1hour}{60minutes}=2.75hours$

Writing the equation:

$264miles=(2x+14mph)*t\\\\264miles=(2x+14mph)*2.75hours\\\\2x+14mph=\frac{264miles}{2.75hours}\\\\2x=96mph-14mph\\\\x=\frac{82mph}{2}=41mph$

So, the speed of the first car is equal to 41 mph.

Now, for the second car we have that:

$SecondCarSpeed=FirstCarSpeed+14mph\\\\SecondCarSpeed=41mph+14mph=55mph$

We have that the speed of the second car is equal to 55 mph.

Hence, the answers are:

$FirstCarSpeed=41mph\\SecondCarSpeed=55mph$

Have a nice day!