Two cars leave towns 760 kilometers apart at the same time and travel toward each other. one car's rate is 18 kilometers per hour less than the other's. if they meet in 4 hours, what is the rate of the slower car?

Respuesta :

Let s be the speed of the first car; and s-18 be the speed of the other. Then:
760/(s+s-18)=4
760=8s-72
8s=832
s=104
The first car goes 104 kph; the second car goes 86 kph
☺☺☺☺

The speed of the slower car is 86 kilometers per hour.

What is speed?

The speed is the rate of change of position of an object in any direction. Speed tells us how fast or slow an object travels and describes the distance traveled divided by the time taken to cover the distance.

For the given situation,

The distance between the two cars, d = 760 kilometers

The speed of one car's rate is 18 kilometers per hour less than the other's.

Let us consider the speed be x.

Speed of one car = x,

Then speed of another car = x - 18

Time taken to meet each other, t = 4 hours.

The two cars are traveling in the opposite direction, so we need to add the two speeds of the car.

Total speed = [tex]x+(x-18)[/tex]

The formula of speed is

[tex]Speed = \frac{Distance }{Time}[/tex]

⇒ [tex]Time = \frac{Distance }{speed}[/tex]

⇒ [tex]4=\frac{760}{x+x-18}[/tex]

⇒ [tex]4=\frac{760}{2x-18}[/tex]

⇒ [tex]8x-72=760[/tex]

⇒ [tex]8x=832[/tex]

⇒ [tex]x=104[/tex]

Thus, the speed of the slower car is

⇒ [tex]x-18=104-18[/tex]

⇒ [tex]86[/tex]

Hence we can conclude that the speed of the slower car is 86 kilometers per hour.

Learn more about speed, distance and time here

https://brainly.com/question/2004627

#SPJ2