MHB Find Acceleration: Car Stops from 100 m/s in 410 m Distance

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To find the acceleration of a car that stops from 100 m/s over a distance of 410 m, the correct formula to use is a = (v_f^2 - v_i^2) / (2 * Δx). The final velocity (v_f) is 0 m/s, and the initial velocity (v_i) is 100 m/s, with Δx being 410 m. Substituting these values yields an acceleration of -12.2 m/s², indicating deceleration. The discussion clarifies that negative acceleration is valid and emphasizes the importance of using the correct equation and values. Understanding these concepts is crucial for solving similar physics problems accurately.
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Find the acceleration of a car that comes to a stop from a velocity of 100 m/s in a distance of 410 m.Answer: -24.39 The answer has been given back to me as incorrect

is this because I inputed a negative value and acceleration cannot be negative?Thanks!
 
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Hi mathkid3, :)

Acceleration can be negative. Think of it as deceleration. So that's not the issue.

What equation did you use to solve for "a"? I suggest using ${v_f}^2={v_i}^2+2ad$
 
If we assume constant acceleration $a$, then we note that:

(1) $\displaystyle \Delta v=v_f-v_i=at$

(2) $\displaystyle \Delta x=\frac{1}{2}at^2+v_it$

From (1), we may state:

$\displaystyle t=\frac{v_f-v_i}{a}$ and substituting into (2) we find:

$\displaystyle \Delta x=\frac{1}{2}a\left(\frac{v_f-v_i}{a} \right)^2+v_i\left(\frac{v_f-v_i}{a} \right)$

Multiplying through by $2a$ we obtain:

$\displaystyle 2a\Delta x=v_f^2-v_i^2$

This is equivalent to the relation cited by Jameson. And so:

$\displaystyle a=\frac{v_f^2-v_i^2}{2\Delta x}$

Now plug-n-chug! :cool:
 
a = (0-85) / 2deltax

what is delta x ?

I can only get to the above and/or if I ignore delta x I would get something like acceleration = -42.5 or deceleration is this casehelp Mark!
 
$\Delta x$ is how far the car moved during its acceleration, which is given as 410 m.

$v_f$ is the final velocity, and since the car came to a stop, this is $0\,\dfrac{\text{m}}{\text{s}}$

$v_i$ is the initial velocity which is given as $100\,\dfrac{\text{m}}{\text{s}}$

So, plug in those values (along with the units, it is important in a physics course to get used to carrying the units too) and what do you get?
 
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