For my physics class, I have to calculate an objects acceleration while it came to a stop. It was traveling at $6.26\: \mathrm{\frac{m}{s}}$ at impact, and travels a distance of $0.025\: \mathrm{m}$ while stopping. I assumed I would use a simple kinematics equation to solve. $$V_f^2 = V_i^2 + 2a\varDelta d$$
$$0 = 6.26^2 + 2*a*.025$$ $$-39.2 = .05a$$ $$a = -784\: \mathrm{\frac{m}{s^2}}$$ Is that right? It seems extremely high.
Well, let's think about this: An object is traveling at 6.26 m/s during impact and travels only 0.025 meters before stopping. The force causing the object to decelerate needs to be extremely high. It's just like having a force being applied for a very short period of time, such as a bat hitting a baseball and the time of contact is extremely small, you would, as a result, get an extremely high answer. So yes, your answer does make sense :-)
This is correct, assuming constant acceleration, we have for this problem
$$a = -\frac{v^2_i}{2d} = -\frac{(6.26)^2}{0.05} = -784\frac{m}{s^2}$$
First, I applaud you for asking the question. Too often, I have graded homework and tests where numbers were submitted for answers without any thought as to whether they were reasonable.
However, this is a collision and the acceleration, though brief, can be large as is this one, about $-86g$.
For examples, see this hyperphysics article on car crash decelerations.
