Physics Acceleration Problem

[mathkid3]

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!

 

[Jameson]

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 ${v_f}^2={v_i}^2+2ad$

 

[MarkFL]

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!

 

[mathkid3]

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 case


help Mark!

 

[MarkFL]

$\Delta x$ 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?