Question

help? a man drives his car at a uniform rate from his home toward albany ny. the man;s son left the same house 45 minutes later than his father and traveled along the same road at a rate 12miles per hour faster than his father. the son overtook his father 130 miles from home. what was the rate at which the father drove?

Answer 1

lool. I'm doing the same type of question. I'm stuck too

Answer 2

lol

Answer 3

@riccyscaduto

Answer 4

anyone know what to do? lol

Answer 5

kinza

Answer 6

I should be able to do this eventually, but I'm not your best bet for the right answer. :*

Answer 7

how do i tag someone?

Answer 8

well i know the answer is either 40 or 2.4 haha

Answer 9

@ and then username

Answer 10

its multiple choice , but i would like to know how to actually do the problem

Answer 11

@agent0smith

Answer 12

I have an way of getting 40, but it's a little messy

Answer 13

please explain =)

Answer 14

Give me two minutes to see if there's a neater way of doing it

Answer 15

ok, take your time

Answer 16

Got it. :D

Answer 17

:D

Answer 18

If you let x be the speed that the father is travelling in miles per hour
and t be the time travelled in miles/hour,
the equation for the distance the father travels is:

\[xt\]
and for the son it's:

\[(x+12)(t-\frac{ 3 }{ 4 })\]

Answer 19

Because the son travels 12 miles/hour faster than his father, and he takes 45 minutes less to reach that same distance.

You have to change 45 minutes to hours, because the speeds are in miles/hour.

Answer 20

Are you following, Mr. Noodles?

Answer 21

I'll continue when you are. qq

Answer 22

ok, i get this so far

Answer 23

Okay, so those are equations for the distances travelled by both father and son, right?

Answer 24

yes

Answer 25

We know that the son overtakes his father 130 miles from home--after a distance of 130 miles. So let both equations equal 130.\[xt = 130\]\[(x+12)(t-\frac{ 3 }{ 4 })=130\]
and now you have a system of equations you can use to solve for x, which we initially established as the speed the father was travelling.

Answer 26

once u solve for x?

Answer 27

wont it be in terms of t?

Answer 28

=(

Answer 29

You're solving for t first to find x, right? Because in the equations we've created here, the time taken should be equal--we've taken the 45 minute delay of the son into consideration numerically.

Answer 30

so u would set xt=(x+12)(t-3/4) equal to eachother?

Answer 31

Not how I did it. Solve for one variable first--you can't work with two.

Answer 32

Solve for t first.

Answer 33

t=130/x

Answer 34

then plug it in?

Answer 35

Oh, that could work, yeah.

Answer 36

is that how you did it?

Answer 37

Not exactly, but that method works just fine. =)

Answer 38

Once you end up with a quadratic, tell me what it is and I can confirm it.

Answer 39

so plugging it in would get you x(130/x)=130?

Answer 40

or am i not doing this right?

Answer 41

Haha, you're plugging it into the wrong equation. Why not solve it my way?
Expand both equations and isolate for t, then let t=t and solve for x.

Answer 42

ok that seems easier lol

Answer 43

so 130/x=(130/x+12)+3/4

Answer 44

Hm, not quite.

Answer 45

\[xt = 130\]\[t = \frac{ 130 }{ x }\]

\[(x+12)(t-\frac{ 3 }{ 4 })=130\]\[xt -\frac{ 3x }{ 4 }+12t -9=130\]\[(x+12)t-\frac{ 3x }{ 4 }=139\]\[t=\frac{ 139+\frac{ 3x }{ 4 } }{ x+12 }\]

Answer 46

ok so then what would u do to make it a quadratic?

Answer 47

Now we know what t is equal to...so t = t and rearrange to create a quadratic to solve for x.

Answer 48

i did not get an even answer, how do u rearange it to a quadratic? i did not do it correctly

Answer 49

\[t=t\]\[\frac{ 130 }{ x }=\frac{ 139+\frac{ 3x }{ 4 } }{ x+12 }\]\[(x+12)(130)=(139+\frac{ 3x }{ 4 })(x)\]\[130x+1560=139x+\frac{ 3x^{2} }{ 4 }\]\[0=\frac{ 3x^{2} }{ 4 }+9x-1560\]

Use the quadratic equation to solve for x.

Answer 50

oh ok, so u get -52 and 40 nd have to reject the negative?

Answer 51

thank you so much!!! how do i give u medals u deserve like 100 lol ?

Answer 52

Hahahaha, you're very welcome. And of course, his speed can't be negative. Speed isn't a vector, it has no direction, so it's always positive.

Answer 53

thank you again, finally i can sleep! have a good night =)