accelerationhomework-and-exerciseskinematics
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.
When stopping the engines, how quickly will a ship lose its speed, and how far will it go?
Newton's law tells us the change in the ship's momentum equals the drag force:
$$M \frac{dv}{dt} = - F_{drag}$$
Here $M$ is the ship's mass, and $v$ is its speed. For ships with a large areal cross section $A$ under the water line and a speed $v$ such that $\sqrt{v^2 A} >> \nu$ with $\nu$ the kinematic viscosity (momentum diffusion constant) of the water, the drag force is given by:
$$F_{drag}= \frac{1}{2} C_D \rho v^2 A$$
Here, $\rho$ is the density of the water, and $C_D$ the drag coefficient, a dimensionless constant typically in the range 0.1 - 0.5, depending on the shape of the ship.
This is all you need. The rest is straightforward math. Substituting the equation for the drag force into Newton's law, one readily obtains
$$\frac{dv}{dt}= \frac{-1}{L}v^2$$
With $\frac{1}{L} = \frac{C_D \rho A}{2M}$. The solution to this equation is $v = L/(t+t_0)$ with $t_0$ chosen such that the ratio $L/t_0$ matches the initial speed of the ship.
Clearly, although the ship will shed its speed rapidly at early times, at later times the speed loss slows down considerably. The distance travelled is the integral over $v(t)$:
$$x(t) = L \ln{\frac{t+t_0}{t_0}}$$
Some specific results:
If it takes a time $t_0$ and a distance $(\ln 2) L \ = \ 0.693 L$ to half the ship's speed, it will take an additional time $2t_0$ and an additional distance $0.693 L$ to again half the speed. The total time to reduce the speed by 90% is $9t_0$. During that time period the ship will travel a distance of $2.30 L$
Estimation of the parameter $L$ and $t_0$ from velocity vs time data is easy: $t_0$ is the time it takes to reduce the initial speed $v_0$ to half the value, and $L_0$ is the product $v_0 t_0$.
Note that the derived results are valid up to times $t$ at which $v(t)\sqrt{A} >> \nu$ or $t+t_0 << L \sqrt{A}/\nu$.
You are right in that gravity did not change during data collection. You are a victim of uncertainty, which is a very important part of experimental physics. I'm sorry in advance for the "wall of text", and I hope that this clears up some confusion.
The problem is that $1.50$ may not be exactly $1.500000000...$. Because the numbers are provided rounded, they are not exact and you have lost information. Imagine that my watch can only report time to the nearest second and my measuring tape only reports to the nearest meter. If I measure a car's movement and it moves 2.3 meters in 0.8 seconds (true measurements, 2.875 m/s) I am required to round all my data to the nearest round number (2 meters in 1 second, 2.0 m/s). So, if I calculate a number based on my rounded data, it won't perfectly reflect reality because my data do not perfectly reflect reality. Even though your numbers are more precise than my example (accurate down to hundredths of a second and centimeters), there is still some amount of uncertainty in the numbers.
Feel free to skip to the "What it all means" section. The stuff after this is pretty dry and obtuse.
Quantitative Explanation
Note that I'm going to call displacement $x$ and time $t$, just for convenience.
Let's have a look at a subset of the data as an example:
Time (s) | Displacement (m)
1.50 | 3.09
2.00 | 6.60
You know the displacement to two decimal places-- that means that, for $t = 1.50$, $x$ could be anywhere from $3.085$ to $3.094999... \approx 3.095$, and it would still be okay to call it $3.09$, as long as you're rounding to that number of significant digits. Similarly, the time might not be exactly $1.5$! So, if you calculate anything based on those numbers, there's a certain amount of uncertainty in the result. Since there are two variables ($t$ and $x$), they can both vary at once. With the equation $v = (x_f - x_i)/ \Delta t = $, $v$ is biggest when $\Delta x$ is maximized and $\Delta t$ is minimized (dividing by a smaller number yields a larger number), and $v$ is smallest when $\Delta x$ is smallest and $\Delta t$ is biggest. We don't know the true, unrounded values of the two variables, so any possible combination of them is just as good as any other. Here's the range of possible velocities based on the above displacements:
$t = 1.5 \rightarrow 2.0$:
$$ v_{max} = \frac{6.605 - 3.085}{1.995 - 1.505} \approx 7.18 \, \mathrm{m/s} $$
$$ v_{mid} = \frac{6.600 - 3.090}{2.000 - 1.500} = 7.02 \, \mathrm{m/s} $$
$$ v_{min} = \frac{6.595 - 3.095}{2.005 - 1.495} \approx 6.86 \, \mathrm{m/s} $$
Quite a range! Let's do it again for the next time so that we can get a range of accelerations:
$t = 2.0 \rightarrow 2.5$:
$$ v_{max} = \frac{9.695 - 6.595}{2.495 - 2.005} \approx 6.36\, \mathrm{m/s} $$
$$ v_{mid} = \frac{9.690 - 6.600}{2.500 - 2.000} = 6.18 \, \mathrm{m/s} $$
$$ v_{min} = \frac{9.685 - 6.605}{2.505 - 1.995} \approx 6.04 \, \mathrm{m/s} $$
Again, quite a range. Now, to find the max,min accelerations possible for your data, you take the same approach. For $a_{max}$, divide the biggest possible $\Delta v$ by the smallest possible $\Delta t$, and the reverse for the min. For the above numbers, you get accelerations like so:
$$ a_{min} = \frac{6.36 - 6.86}{2.505 - 1.995} \approx -0.98 \, \mathrm{m/s}^2 $$
$$ a_{mid} = \frac{6.18 - 7.02}{2.5 - 2} = -1.68 \, \mathrm{m/s}^2 $$
$$ a_{max} = \frac{6.04 - 7.18}{2.495 - 2.005} \approx -2.33 \, \mathrm{m/s}^2 $$
where "min" and "max" are used (somewhat sloppily) to mean "greatest in magnitude", not the true meanings of "minimum" and "maximum". If I haven't messed up the numbers, the intermediate accelerations could be anywhere in the above range!
What it all means: Now you should be able to recognize that, because your accelerations are derived from numbers of limited certainty and they fall within the range given above, they cannot really be said to be different-- they are said to be "within uncertainty" of each other. It would be appropriate to take the average of all your values and call that the best guess you have of the acceleration. If you wanted to be a true scientist about it, you might have a go at calculating the uncertainty in acceleration, which is nontrivial, or you could go the lazy route (like a lot of scientists :)) and use something like the standard error of all your acceleration values. Either way, the end-all is that the acceleration didn't change during the data collection, it only looks like it did because you didn't measure position and time precisely enough :)
If you're interested in reducing the uncertainty, read on!
These numbers are not acceptable, the range is too big. The problem is that our uncertainties ($0.005 \mathrm{m}$ and $0.005 \mathrm{s}$) are very large compared to our numbers: we're using time steps of half a second, but we have an uncertainty of $2 \cdot 0.005 = 0.01$ in each variable (twice the uncertainty because you're taking the difference of two values). You can reduce the effect by evaluating the difference across more than one time step (ie find the acceleration from $t=0 \rightarrow 1$ instead of $0 \rightarrow 0.5$. This way, the change in time (and displacement) is bigger, but the uncertainty remains the same, so it impacts the result less!
Best Answer
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 :-)