Two cars, 127 miles apart, start moving towards each other at the same time. One is moving 3 times as fast as the other. If they meet 1.5 hrs later, find the average speed of the slower car in miles per hour?

slower car moves 127/4 miles in 1.5 hrs

(127/4 mi) / 1.5hr = 21.17 mi/hr

Sorry, how did you get that? :(

v1•t+v2•t=s

v1=v, v2=3v,
v•t+3v•t=4•v•t=s
v=s/4•t =127/4•1.5 =21.17 mph

Oh! I understand now, thanks so much to both :)

To find the average speed of the slower car, let's first find the distances traveled by both cars.

Let's assume the speed of the slower car is 'x' miles per hour. Since the other car is moving 3 times as fast, its speed would be 3x miles per hour.

Now, we know that speed equals distance divided by time. So, we can use the formula:

Speed = Distance / Time

For the slower car:
Distance traveled by the slower car = Speed * Time
Distance = x * 1.5

For the faster car:
Distance traveled by the faster car = Speed * Time
Distance = (3x) * 1.5

Since the two cars are moving towards each other and their meeting point is 127 miles apart, the sum of their distances traveled should be equal to 127 miles.

So, the equation becomes:
x * 1.5 + (3x) * 1.5 = 127

Now, let's solve this equation to find the value of 'x'.

x * 1.5 + 4.5x = 127
5.5x = 127
x = 127 / 5.5

Now, let's calculate the value of 'x':

x = 23.09090909090909

So, the average speed of the slower car is approximately 23.09 miles per hour.