Question

two trains start 5 min apart. Starting from rest, each is capable of a maximum speed of 300 km/h after uniformly accelerating over a distance of 4 km. (a) What is the acceleration of each train? (b) how far is the first train ahead when the second starts? (c) How far apart are they when they are both traveling at maximum speed?

Answer 1

I can only use these equations:
\[v=v_{0}+at\]
\[r=r_{0}+v_{0}t+\frac{1}{2}at^{2}\]

Answer 2

i really should be quiet because i have no idea how to do this, but after reading it 3 times it strikes me that something is missing
how in the world are you supposed to know the acceleration?

Answer 3

I'm not sure haha. I'm trying to rack my brain for something...

Answer 4

are you sure there is not some missing part ? like "the meet one hour later" or something?

Answer 5

i have two cars
they do not accelerate at the same rate
if i ask you for the rate, you cannot tell me

Answer 6

No I wrote it word for word from my work and it tends to be right so I'm not sure.

Answer 7

ooh maybe this line
tarting from rest, each is capable of a maximum speed of 300 km/h after uniformly accelerating over a distance of 4 km.
means that it takes 4 km to accelerate to a speed of 300

Answer 8

\[v=v_{0}+at\] if the start at rest then it is
\[v=at\]

Answer 9

yeah I'm trying to work with that but I keeps messing up I have the answer:
\[0.87 m/s^{2}\]

Answer 10

i have an idea (could be wrong)

Answer 11

\(v_0=0\) since they start at rest and also \(r_0=0\)

Answer 12

then \[300=at\]and
\[\frac{1}{2}at^2=4\]

Answer 13

we can solve that for \(t\) then for \(a\)

Answer 14

\[\frac{.5at^2}{at}=\frac{4}{300}\implies t=\frac{8}{300}\]

Answer 15

no that doesn't seem to work, damn

Answer 16

maybe start by doing this?
\[4km*\frac{1000m}{1km}=4000m\]
\[\frac{300km}{h}*\frac{1000m}{1km}=\frac{300000m}{h}\]

Answer 17

everything is given in terms of km i would keep those units until the end
but i did not notice that your answer was in terms of meter, so lets see if what i wrote is correct

Answer 18

nope it doesn't even make sense

Answer 19

aww man lol

Answer 20

isn't the internet wonderful? i wish they had this when i was in school
https://answers.yahoo.com/question/index?qid=20110125211241AAqffOy

Answer 21

is there a way to solve for one equation and put it into the other

Answer 22

or here
https://answers.yahoo.com/question/index?qid=20110125171621AACjcmk

Answer 23

awesome thank you!

Answer 24

yw