at 8:00am the smiths left a campground, driving at 48 mi/h. at 8:20am the garcias left the same camp ground and followed the same route, driving at 60 mi/h. at what time did they overtake the smith's?
How far did the Smiths drive from 8 a.m. to 8:20 a.m.; i.e., in 20 minutes?
distance = rate*time.
d = 48mi/hr * 1/3 hr = 16 miles which the Smiths had as a head start.
let x = distance driven by Garcias. time = d/r = x/60.
time for Smiths = x-16 is distance driven. time = (x-16)/48.
Set time = time or
x/60 = (x-16)/48
solve for x.
That will be the distance. Plug that into x/60=time and calculate time. Add that to 8:20 to know the final time at which both reached the same point. Post you work if you run into trouble.
To determine the time at which the Garcias overtook the Smiths, we need to find the distance driven by the Garcias.
We know that the Smiths had a head start of 16 miles because they left at 8:00 am and drove for 20 minutes at a speed of 48 miles per hour.
Let x be the distance driven by the Garcias. Since the Garcias and the Smiths followed the same route, the time it takes for the Garcias to travel that distance is x/60, as they were driving at a speed of 60 miles per hour.
The time it takes for the Smiths to travel the same distance is (x-16)/48, as they were driving at a speed of 48 miles per hour.
Now we set up the equation and solve for x:
x/60 = (x-16)/48
To solve for x, we can cross-multiply:
48x = 60(x-16)
48x = 60x - 960
960 = 12x
x = 80
So, the Garcias drove a distance of 80 miles.
To calculate the time it took for the Garcias to overtake the Smiths, we can plug this distance into the equation x/60 = time:
80/60 = 4/3 hours
To convert this to minutes, we multiply by 60:
4/3 * 60 = 80 minutes
So, the Garcias overtook the Smiths after driving for 80 minutes.
To find the time at which they overtook the Smiths, we add 80 minutes to the Garcias' departure time of 8:20 am:
8:20 am + 1 hour, 20 minutes = 9:40 am
Therefore, the Garcias overtook the Smiths at 9:40 am.