Solve Final Velocity & Time for Train Traveling 200m

[tcharger130]

Homework Statement

A train on a straight, level track has an initial speed of 35.0 km/h. A uniform acceleration of 1.50 m/s sq is applied while the train travels 200 m. (a) What is the speed of the train at the end of the distance? (b) How long did it take for the train to travel the 200m?

Homework Equations

?

The Attempt at a Solution

I dropped into this physics class late and I don't know how to properly derive the formula I think I need. I converted 35km/h to 14m/s but I don't know a formula to use now.

Thanks

 

[CompuChip]

The two formulas for a motion with uniform acceleration (a = constant) are
$\Delta v = a \Delta t$
and
$\Delta s = \tfrac12 a (\Delta t)^2$

where $\Delta v$ and $\Delta s$ are the change in velocity and covered distance, respectively, during a time $\Delta t$.

 

[tcharger130]

but i don't have time?

 

[thebigstar25]

your start is good, you decided to convert the speed from km/h to m/s,im not sure how you did it but it was not correct, to convert from km/h to m/s just multiply that number with (1000/3600) to get it in m/s..

For first part :

It is asking for final velocity and you are given the initial velocity , the acceleration and the distance , you can use this equation:
Vf^2 = Vo^2 + 2ax


Then when you get Vf you can easily get the time using this equation:

Vf = Vo + at

 

[CompuChip]

but i don't have time?

No, but you can calculate it.
Look what you've got, you are given $\Delta s$ and a. That allows you to use the second formula to find $\Delta t$. Then you can use that result in the first formula to calculate the increase in velocity.

Of course, you can also directly use the formulas given by bigstar, which are derived from the two I gave. Personally I prefer doing it my way, because I need to remember less equations :)[edit]
Actually the second formula should be
$$\Delta s = \frac12 a (\Delta t)^2 + v_0 \Delta t$$
where v₀ is the initial velocity. The one I gave earlier applies for v₀ = 0, i.e. when you accelerate from rest (which is not the case here).