A slow train which stops at every station passes a certain signal box at noon.

[Kitimbo]

A slow train which stops at every station passes a certain signal box at noon.Its motion between the two adjacent stations is such that it is s km past the signal box t min past noon,where s=⅓t+¹/9t²-¹/27t³.Find the time of departure from the first station,and the time of arrival at the second.

So wat I did I made s=0 so I solved 0=1/3t+1/9t²-1/27t³.
And I got t=0 and t =-3(-1+√5)/2 and t=-3(-1-√5)/2.
But the answer is 11.59am ,12.02pm.
I think am wrong

 

[JeffM]

A slow train which stops at every station passes a certain signal box at noon.Its motion between the two adjacent stations is such that it is s km past the signal box t min past noon,where s=⅓t+¹/9t²-¹/27t³.Find the time of departure from the first station,and the time of arrival at the second.

So wat I did I made s=0 so I solved 0=1/3t+1/9t²-1/27t³.
And I got t=0 and t =-3(-1+√5)/2 and t=-3(-1-√5)/2.
But the answer is 11.59am ,12.02pm.
I think am wrong

I don't like the wording of the problem, but it appears to be a differentiation problem.

\(\displaystyle t = time,\ 0 = noon.\)

\(\displaystyle s = distance\ from\ signal\ box,\ s < 0 \implies distance\ toward\ first\ station.\)

\(\displaystyle s = f(t) = \dfrac{t}{3} + \dfrac{t^2}{9} - \dfrac{t^3}{27}.\)

\(\displaystyle f'(t) = 0 \implies \dfrac{1}{3} + \dfrac{2t}{9} - \dfrac{3t^2}{27} = 0 \implies\)

\(\displaystyle t^2 - 2t - 3 = 0 \implies t = 3\ or\ t = -\ 1.\)

noon - 1 minute = 11:59.

noon + 3 minutes = 12:03.

I don't get 12:02, but I am not seeing my error. Maybe the book has a typo.

 

[ThomasTrig]

Yeah . . .

I don't get the 12:02 part either. @Kitimbo - Can you double check that on your end. Thx!