math help please???

Question

At 1:00 p.m., a car leaves a city and travels north at a rate of 55 mi/h. An hour later, a second car leaves the city and travels south at a rate of 60 mi/hr. At what time will the two cars be 285 miles apart?

Answer 1

Answer: 4PM

South<=======Start======>North
4PM===3PM===1PM==3PM===4PM
120 M==60 M===0 M==110 M==165 M

120 Miles +165 Miles = 285 Miles

Answer 2

Every hour, the cars become 115 miles further apart. Since the first car has a 1 hour headstart, they are 55 miles apart after 1 hour.

After 2 hours, the 1st car will have gone 110 miles, while the second will have gone 60. They are then 170 miles apart.

After 3 hours, the 1st car will have gone 165 miles, while the second will have gone 120. They are then 285 miles apart.

Therefore they will be 285 miles apart 3 hours after the first car leaves the city, or 4:00 p.m.

Answer 3

Let t be the hours traveled by the first car. Then (t-1) = the hours traveled by the second car since it left an hour later.
Distance = speed times time, so 285 = 55t + 60(t-1)
You add the distances of each car since they are going in the opposite direction.

This is 285 = 55t + 60t - 60, or 285 = 115t - 60

Add 60 to both sides then divide both by 115 to get t; add this to 1:00 to get the time they are 285 miles apart.

Answer 4

after 1 hr, they are 55 miles apart. they are then moving apart at 55+60=115 mph.
the distance will be d=55+115t where d is the distance between the cars & t is the time from 2pm
285=55+115t
115t=285-55=230
y=230/115=2 hrs

This will occur at 4pm

Answer 5

D=RT
285=55(x)+60(x-1)
285=55x+60x-60
345=115x
x=3

1:00 pm plus an additional three hours equals 4:00 pm

Answer 6

285-55=230
55+60=115
230÷115=2
1:00 p.m+1:00+2:00=4:00 p.m

Answer 7

We add the speeds of the cars when they are both in motion but since the 55mph car takes off 1 hour before, we must add that to the equation.

(60+55)x+55 = 285
115x+55=285
115x=230
x=2 hours
+1 hour for the 55 miles we added for the first car.
So the answer in 4 pm.