help with word problems...?

Question

i'm not good at word problems and i could use some help!!
for certain problems i can't figure out what to write as the equation... i know how to solve the equation but i don't know what to wrte for it:

a passenger train leaves south station 2 hours after a freight train leave the same station. the freight train is treveling 20 mph slower than the passenger train. find the rate of each train if the passenger train catches up with the freight train in 3 hours.

equation??

Answer 1

let x be rate of speed of passenger train
and then (x-20) is the speed of the freight train -- you see it is slower?
so if the passenger train is traveling for 3 hours to catch up to the freight train and it left 2 hours after the freight train that means the freight train will have travelled for 5 hours when the passenger train catches up to it.

the formula d=rt means
distance equals rate of speed multiply time
and when the PT catches up to the FT they will have travelled the same distance.

write 3*x = 5(x-20) and solve
remember x is the speed of the PT

Answer 2

Let x be the speed of the freight train
the distance traveled in 2 hour = 2x

Relative speed of the two train = 20
distance travel for 3 hr = 60

2x = 60. X = 30
freight train travels 30 mph and passenger train travels 50mph

Check. Freight train travels 5 hr = 30x5 = 150 mi
passenger train travels 3 hr = 50 x 3 = 150 mi

Answer 3

The question is basically say;
It will take 3 hours for one train to catch up to another train, if the slow train gets a 2 hour head start and (heres the important part) the speed at which they get closer and closer is 20mph
We will need to calculate the individual speeds, and we know the distance they travel is the same, so the relationship between the speed is: 3x = 5y (the passenger train will go 3 times(in miles) its speed(per hour) in the time traveled(3hours), and the freight train will go 5 times(in miles) its speed(per hour) in 5 hours.
Re-writing the speed equations we get:
x = 5y/3 ... & ... y = (3/5)x
nwo you can use S = d/t to work out the specific values (remember to add the +2hours, for the earlier departure)

Answer 4

let the rate of passenger train = x mph

so the rate of freight train = (x - 20) mph

the passenger travelled for 3 hrs to catch up the freight train

the freight train travelled for 5 (3+2) hrs in the mean time

the distance travelled by passenger train in 3 h = 3*x m

the distance travelled by freight train in 5 h = 5(x-20)

when they meet they travelled same distance because starting point is same.

3x = 5(x-20)

3x = 5x - 100

2x = 100

x = 100/2 = 50 mph

x-20 = 50 - 20 = 30 mph

so the rate of passenger train = 50 mph and

the rate of freight train = 30 mph