Train Velocity Model for Distance of 5280 ft

[member 428835]

Homework Statement

A piecewise-continuous function models the velocity of a train (fps) as follows:
$$
v(t) =
\begin{cases}
80, & \text{if }\text{t>15} \\
bt+ct^2+dt^3, & \text{if }t\text{ <15}
\end{cases}
$$

How much time does it take the train to travel 5280 ft?

Homework Equations

$$(1) vdv=ads$$
$$(2) ds=vdt$$
$$(3) dv=adt$$

where s is distance, v is velocity, t is time, a is acceleration

The Attempt at a Solution

i have: $5280=80t+15^2b+15^3c+15^4d$ by integrating the piecewise function and using $(2)$. it seems if i can find variables $b,c,d$ i will be done.

i have $80=bt+ct^2+dt^3$ from the continuity of the function (given)

by $(3)$ i have $0=b+2*15c+3*15^2d$

it seems if i can find one more equation for $b,c,d$ i'll be done. i don't think i can use $(1)$ as i do not have velocity as a function of distance. I've used the other two equations, so i feel i am close. any help is very appreciated. Thanks!

 

[mfb]

Your equation 5280=... looks wrong. First, 80t is not the distance traveled after the first 15 seconds. Second, check the integration itself. Third, at that point you cannot know if the 5280 ft will be reached in the first 15 second or not.

i have $80=bt+ct^2+dt^3$ from the continuity of the function (given)

This is not the equation you have. Where (at which time!) are the functions equal?

Is the function piecewise-continuous (which does not help) or piecewise-continuous differentiable?
You could add the requirement that the second derivative is the same at both sides as well, but I don't see this requirement in the problem statement.