I get two different answers when solving a kinematics problem

[LugubriuousLamia]

1. A car accelerates for 200 meters and has an initial velocity of 2.0 m/s and a final velocity of 6 m/s. What is the acceleration of the car if this change in velocity takes 12 seconds?

Homework Equations

a=Δv/Δt
Δx=Vi*t+.5at²[/B]

The Attempt at a Solution

When I solve this using the first equation I get the acceleration is equal to .33 m/s² because (6m/s - 2m/s)/ 12 seconds = .33 m/s²

However when I use Δx=Vi*t+.5at² I get[/B]

200m= 2m/s*12 sec+.5*a*12 sec<sup>2
With a being equal to 2.4 m/s2

Which of these answers would be the correct answer to this problem. I am inclined to believe that the correct answer would be the first answer. However I am not sure as to why this would be the case. I am looking for a concrete reason as to why one solution would be more correct than the other.</sup>

 

[Orodruin]

Assuming that the acceleration is constant, your results show that the assumptions on the distance, velocities, and time are incompatible. Note that (12 s)(6 m/s) = 72 m so even if the car was moving at the listed final speed for all of the 12 s, it would not reach a distance of 200 m.

 

[DrClaude]

The only thing I can see is that the equation based on $\Delta x$ assumes that acceleration is constant, which is not specified in the problem. The best you can do is to calculate the average acceleration, using $\Delta v / \Delta t$.

Edit: beaten by @Orodruin

 

[PeroK]

1. A car accelerates for 200 meters and has an initial velocity of 2.0 m/s and a final velocity of 6 m/s. What is the acceleration of the car if this change in velocity takes 12 seconds?

For future reference: if you have uniform acceleration from $2m/s$ to $6m/s$, then the average velocity during this time is $4m/s$.

And, if the time for this acceleration is $12s$, then the displacement during this time is $\Delta x = (4m/s)12s = 48m$.

Note: In general, the average velocity is $\frac12 (v_i + v_f)$ (assuming constant acceleration). Which is, in fact, half way between the two.

Also, if a car travels $200m$ in $12s$, then its average velocity during this time is $16.7m/s$, which is significantly greater than the speeds involved in your question.

Using average velocity is perhaps a good way to get a bit more intuition about the sort of numbers you expect in answer to these questions.

 

[LugubriuousLamia]

Thank you for the replies, it would seem that the problem is inherently flawed so there is no actual answer than can be obtained.

 

[Orodruin]

That depends on your definition of "inherently flawed". I could certainly produce an acceleration profile that satisfies the given data, but it will not be a motion with constant acceleration.