Not sure why I can't use v=v0 + at on this problem.

[remcalister]

Homework Statement

Freight trains can produce only relatively small acceleration and decelerations.

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0400 m/s^2 for 8.00 min, starting with an initial velocity of 4.00 m/s?

Homework Equations

v=v0 + at
x=x0 +v0t +1/2at^2
v^2=v0^2 +2a(x-x0)

The Attempt at a Solution

Why can't i use the first equation I listed to solve this problem? Using that equation, you end up with v = 76.8 m/s

If you first use the second equation, and then the third equation, you end up with v = 23.2 m/s

Thanks in advance!

 

[billy_joule]

The first eqn gives the same as the 2nd and third.

v = 76.8 m/s is wrong, 23.2 m/s is right you may have pressed a wrong button on your calculator.

v=v0 + at
v = 4 m/s + (0.04 m/s^2)*(8 min * 60s/min) = 23.2 m/s

 

[olivermsun]

The first equation is exactly what you're supposed to use. Using the second and third seems roundabout.

But how are you getting 76.8 m/s with the first equation? Can you show your work?

 

[remcalister]

Oh goodness, it's been a long weekend haha. I accidentally set it up as 4 m/s TIMES (0.04m/s^2)(480s).

Thanks guys

 

[HalfLostAndSearching]

As a scientist, it is important to understand the limitations of equations and when they can be applied. In this case, the equation v=v0 + at assumes that the acceleration is constant throughout the entire time period. However, in the problem given, it is stated that freight trains can only produce relatively small acceleration and decelerations. This means that the acceleration may not be constant and could change over time. Therefore, the equation v=v0 + at may not accurately represent the situation and it is necessary to use the more general equations x=x0 +v0t +1/2at^2 and v^2=v0^2 +2a(x-x0) to account for any changes in acceleration. These equations take into account the initial velocity, final velocity, and displacement, making them more appropriate for this problem. It is important to always carefully consider the given information and choose the appropriate equation to use in order to accurately solve a problem.