how do u solve this??

Question

each week we get a problem of the week in math for the weekend, i usually get them after a couple minutes, but i've tried for like a reeeeeally long time and i still cant figure it out. can anyone help me??

Makin' Tracks:
A pedestrain is three-eighths of the way across the bridge when he hears a train coming from behind. If he runs as fast as possible back toward the train, he will get off the bridge just in time to avoid a collision. Also, if he runs as fast as possible away from the train, he will get off the bridge just in time to avoid a collision. the train is traveling at 60 miles per hour. How fast can the person run?

Answer 1

15 miles per hour

Let x = the speed of the person running
Let 8y = the length of the birdge


If the person runs toward the train:

time needed to get off the bridge = 3y/x

at this time, the train just comes to the bridge, so its original distance from the bridge is 60*(3y/x)


If the person runs away from the train:

time needed to get off the bridge = 5y/x

at this time, the train just passes through the bridge, so it travels the length of the birdge + the original distance from the bridge, which is 8y + 60*(3y/x)

since this total distance is also speed multiplies by time, it can also be expressed as 60*(5y/x).

therefore 8y + 60*(3y/x) = 60*(5y/x)

both side can be divided by y:

8 + 60*3/x = 60*5/x

8 = 120/x

x=15

so the answer is 15 miles per hour

Answer 2

Suppose that the length of the bridge is L, that the train is at a distance x from the bridge start and that the unknown velocity of the person is v. Then:

x/60 = 3L/8v (1) (they arrive at the same time at the start of the bridge)

and:

(x + L)/60 = 5L/8v (2) (they arrive at the same time at the end of the bridge)

Subtract (2) - (1) and you have:

L/60 = 2L/8v => 1/60 = 1/4v => v = 15 miles/hour

Answer 3

i <3 boy meets world too

Answer 4

v1=speed of guy
v2=speed of train
t1=time to close side
t2=time to far side
D=distance of bridge
i=initial distance

v2 * t1=i-(v1 * t1)=i-3/8D
v2 * t2=i+(v1 * t2)=i+5/8D

first solving for t1
v2*t1=i-3/8D
t1=(i-3/8D)/v2

v1=(3/8D)/((i-3/8D)/v2)=v2*3/8D/(i-3/8D)

now lets try from the other set of equations first solving for t2
v2*t2=i+5/8D
t2=(i+5/8D)/v2

v1=5/8D/((i+5/8D)/v2)=5/8D*v2/(i+5/8D)

if we had one more variable i or the distance D we could solve for any of the others but we do not. You can at most solve the problem in terms of D or i I would solve it in terms of D

5/8D*v2/(i+5/8D)=v2*3/8D/(i-3/8D)
first solve for i

5/(i+5/8D)=3/(i-3/8D)

5(i-3/8D)=3(i+5/8D)

5i-15/8D=3i+15/8D

2i=30/8D

i=15/8D

now to check if im right both my set of equations should produce the same answer for v1

v1=v2*3/8D/(i-3/8D)
v1=v2*3/8D/(15/8D-3/8D)
v1=v2*3/8D/(3/2D)
v1=v2*1/8D/(1/2D)

v1=v2*5/8D/(i+5/8D)
v1=v2*5/8D/(15/8D+5/8D)
v1=v2*5/8D/(5/2D)
v1=v2*1/8D/(1/2D)

hey look they are the same and the D cancels

v1=v2*1/4

so like the others where saying 15 miles per hour