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I get two different answers when solving a kinematics problem

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?


2. Homework Equations
a=Δv/Δt
Δx=Vi*t+.5at2



3. 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


200m= 2m/s*12 sec+.5*a*12 sec2
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.
 

Orodruin

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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

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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

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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.
 
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

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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.
 

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