Jan 30, 2013 Thread starter #1 M mathkid3 New member Nov 16, 2012 23 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!
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!
Jan 30, 2013 Admin #2 Jameson Administrator Staff member Jan 26, 2012 4,055 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 ${v_f}^2={v_i}^2+2ad$
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 ${v_f}^2={v_i}^2+2ad$
Jan 30, 2013 Admin #3 M MarkFL Administrator Staff member Feb 24, 2012 13,775 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!
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!
Jan 30, 2013 Thread starter #4 M mathkid3 New member Nov 16, 2012 23 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!
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!
Jan 30, 2013 Admin #5 M MarkFL Administrator Staff member Feb 24, 2012 13,775 $\Delta x$ 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?
$\Delta x$ 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?