Should be simple acceleration problem

[JFAN2260]

Homework Statement

Train is traveling between two stations 1000 m apart. Train never reaches cruising speed, instead engineer first accelerates at 0.1 m/s^2 and then immediately brakes at -.5 m/s^2.

How much time is needed to get from one station to the other?
How much of this time is spent accelerating?

Homework Equations

Not really sure. I'm assuming that x=x0 + v0t + 1/2at^2 is used, but I'm not sure how to deal with the two different accelerations and the ratio between the time spent accelerating and braking.

The Attempt at a Solution

The real problem is I'm not really sure where to start. I've had several attempts but can't find a way to relate T1 to T2.

 

[Uniquebum]

We accelerate:
$x_1 = x_0 + v_0t_1 + \frac{1}{2}a_1t_1^2, x_0 = 0, v_0 = 0$
Thus
$x_1 = \frac{1}{2}at^2$

Also we know that
$v_1 = v_0 + a_1t_1, v_0 = 0$
Thus
$v_1 = a_1t_1$


Then we brake:
$x_2 = v_1t_2 - \frac{1}{2}a_2t_2^2$
Thus
$x_2 = (a_1t_1)t_2 - \frac{1}{2}a_2t_2^2$

We know that total distance x = 1000m
Thus
x = x_1 + x_2 = 1000m
But wait we have still two unknown variables, right? $t_1$ and $t_2$
However, we know
$v_1 = a_1t_1$ (as stated before)
and also
$v_2 = v_1 - a_2t_2$ Where $v_2$ is the speed at the end => $v_2 = 0$
Thus
$v_1 = a_2t_2$

And that's all the info you need. Think about it ;).

 

[PhyPsy]

You know that it loses speed 5 times as quickly as it gains speed, so this means that it takes 5 times as long to reach the top speed of the trip as it does to go back down to 0 speed. Thus, 5/6 of the duration of the trip is spent accelerating, and 1/6 is spent decelerating. That should get you started.

 

[JFAN2260]

So does that mean that when something has starting acceleration but no initial velocity that you have to account for the velocity that accrues over time? Like another problem says that a bus is 30 m away and starts accelerating at .600 m/s^2. How fast do you have to run to catch it? The bus has no initial velocity, but would you have to add a factor of velocity when you solve the equation? Originally i had 0=30-vt+.5(.600)t^2 but should I add something to account for the bus's velocity? I'm not trying to sneak homework answers, I don't get it.

 

[Uniquebum]

It depends. I think the real problem with people is actually understanding the equations
$x = x_0 + v_0t + 0.5at^2$
$v = v_0 + at$

You can approach the bus problem in two ways(that i can think of at least):
-Either look at the system as a whole
-Or separate the runner and the bus and look at them as separate systems

Usually in this kind of problems it's easier to just ignore the other part and focus on one at a time.
For the bus
$x = x_0 + v_0t + 0.5at^2$
As you said, the bus has no initial velocity so v_0 = 0. Remember, we're investigating the bus now so let's ignore the runner.
$x = x_0 + 0.5at^2$
Now what is x_0 then? It depends how you look at the system. If you place the origin of the system on the bus in the very beginning, x_0 will be 0. However, if you decide to put the origin on the runner in the beginning, x_0 will be 30m. You can decide whatever origin you wish BUT you need to remember that the origin of the system stays there and you can't move it. You'll calculate all the stuff based on that origin.
Let's put the origin on the runner, thus
$x = 30m + 0.5*0.600m/s^2*t^2$ for the bus.

Now...
While writing i noticed a flaw in the problem. There's no time limit put in which we should catch the bus or limit for the running speed of the runner.
Anyway, i won't go further because of that but all i said before is still valid.

 

[JFAN2260]

Yeah what i realized is that it's not really a flaw, because if you think about it say the runner is walking at like 3 m/sec, then no matter how long he takes, he will never catch the bus becuase it will have increased velocity to an extent that no amount of time can compensate. I think what you're supposed to do is look at the discriminant of the quadratic formula and realize that if the discrimant is below zero, then there is no way that the time will be a real number.