Railway tunnel

24 mph ?

Which way is the train going? Sounds like from behind but please clarify.

The train is going the same way as the man

I think fresco has it.

correctamundo.

Kev, I had to laugh, half the distance in 12, twice in 24. Seems simple enough, on the same day I was sent this nightmare.

A train accelerates at a uniform rate from its starting point. It travels the last one fourth of the distance from the starting point to the finish line in 3 seconds. How long did it take to travel the entire distance from starting point to finish line?

There does not seem anywhere near enough information to form an answer. Can you assist?

Hi Try,

I suppose 45 secs would be too obvious an answer

Second Attempt at acceleration problem

d=kT^2 also .75d = k(T-3)^2

eliminating d gives the the quadratic

T^2 - 24T +36 = 0

T= 22.8 sec appx

Quote Fresco, "T= 22.8 sec appx"

I marvel at your brevity and accuracy. I took a lot longer

Let the total distance travelled be 1.
Let ?'a' be the acceleration.
Let ?'t' be the total time travelled.

Basic calculus tells us that the velocity at time t is a*t and the total distance travelled is a*t2/2.
So we know:
(1) 1=a*t2/2 and
(2) 0.75 = a*(t-3)2/2

Solving both sides for a we have :
(3) 2/t2 = 3/(2*(t-3)2)
4*(t-3)2 = 3*t2
t2 - 24t + 36 = 0
t = (24 +/- 4321/2)/2 (11) t = 6*31/2 + 12

Add 3 for the total distance of 12 + 6*31/2 =~ 22.39230485

Tryagain.

You don't seem to have used the general formula for the solution of a quadratic i.e.

If ax^2 + bx + c= 0,
then x =[ -b +\- sqrt (b^2 - 4ac)]/2a

This may explain the slight variation in our answers.

Regards fresco.

"You don't seem to have used the general formula for the solution of a quadratic i.e.

If ax^2 + bx + c= 0,
then x =[ -b +\- sqrt (b^2 - 4ac)]/2a

This may explain the slight variation in our answers."

It sure do. I have a simple rule; If it don't have pictures, I don't ?'read' it. :wink:

Tryagain,

I made a mistake using the formula !

Your answer is spot on.

"Your answer is spot on."

Which just goes to show, that if I try and tryagain, eventually I will get lucky.