uniform motion problems, i just dont understand it =[?

Question

14) Mr. Fox drove from his home to his factory at the rate of 25 miles per hour and returned by a different route at the rate of 30 miles per hour. the route by which he returned was 5 miles longer then the route by which he went. the return trip took 10 minutes less then the trip out. find the distance Mr. Fox traveled each week?

5) A pilot plans to make a flight lasting 2 hours and 30 minutes. how far can hefly from his base atthe rate of 300 mph and return over the same route at the rate of 200 mph?

7) The rate of a passenger train exceeds the rate of a freight train by 20 mph. It takes the passenger train 1/2 as much time to travel 160 miles as it does the freight train. Find the rate of each train.

Answer 1

14) Mr. Fox drove from his home to his factory at the rate of 25 miles per hour and returned by a different route at the rate of 30 miles per hour. the route by which he returned was 5 miles longer then the route by which he went. the return trip took 10 minutes less then the trip out. find the distance Mr. Fox traveled each week?

Let

r1 = speed to factory
r2 = speed home from factory
d1 = distance of route to factory
d2 = distance of route home from factory
t1 = time to drive to factory
t2 = time to drive home from factory
D = Total distance driven each week

Given

r1 = 25 mi/hr
r2 = 30 mi/hr
d2 = d1 + 5 mi
t2 = t1 - 10 min = t1 - 1/6 hr

Find the total distance T driven each week.
D = 5(d1 + d2)

We have

d1 = (r1)(t1)
d2 = (r2)(t2)

d1 = 25*t1
d2 = 30*t2

Substituting into the second equation:

d1 + 5 = 30(t1 - 1/6) = 30*t1 - 5

Subtracting

5 = 5*t1 - 5
10 = 5*t1
2 = t1
t1 = 2 hours = 120 minutes

d1 = 25*t1 = 25*2 = 50 miles
d2 = d1 + 5 = 50 + 5 = 55 miles

D = 5(d1 + d2) = 5(50 + 55) = 5*105 = 525 miles

5) A pilot plans to make a flight lasting 2 hours and 30 minutes. how far can hefly from his base atthe rate of 300 mph and return over the same route at the rate of 200 mph?

Let

d = one way distance flown
r1 = speed outbound
r2 = speed returning
t1 = time outbound
t1 = time returning

Given

r1 = 300 mi/hr
r2 = 200 mi/hr
t1 + t2 = 2 1/2 = 5/2 hours

Find d.

We have:

t1 = d/r1
t2 = d/r2

t1 + t2 = d/300 + d/200 = 5/2
2d + 3d = 1500
5d = 1500
d = 300 miles

7) The rate of a passenger train exceeds the rate of a freight train by 20 mph. It takes the passenger train 1/2 as much time to travel 160 miles as it does the freight train. Find the rate of each train.

If the passenger train takes half the time to travel a given distance as the freight train, it is going twice as fast.

p = rate passenger train
f = rate freight train

p = f + 20
p = 2f
0 = f - 20
f = 20 mph

p = 2f = 40 mph

Answer 2

The trick is you have to turn the problem into math...

14)

lets designate the distance of the first route A the distance of the 2nd route B

lets call the time to drive route A, T1 and the time to drive route B T2.

So already we can write one equation, since route 2 took 10 minutes less than the first:

T2 = T1 - 10

also since route B is 5 miles longer than route A:

B = A+5

now we have to relate the two, distance travelled = speed x time, we also need to have consistent units: 25 mph = 0.4167 miles/min and 30 mph = 0.5 miles / min

thus: 0.4167 x T1 = A and 0.5 x T2 = B

Now we have 4 variables, and 4 equations, we can solve the problem. Calc A+B that is distance drive each day (if he drives 5 days a week multiple by 5 for your answer).

5) Ok here you know that the distance is the same both ways, lets call the distance X, now you have V1 and V2 which are the 300mph and 200 mph velocities respectively. so:

X = V1 x T1 = V2 x T2 , V1 and V2 are Known, also we know that T1 + T2 = 2.5 hours, so:

300T1 = 200T2
T1+T2 = 2.5

solve for T1 and T2, then you can get X.

7)

VelP = VelF + 20

160 = VelP x TimeP
160 = VelF x TimeF
TimeP = 1/2 Time F

solve...