# Calculating acceleration based on a starting and ending distance

I have been trying to figure this problem out and just can't.

The answer is supposed to be (A): 1.0 m/s/s.

Can anyone tell me how they got that answer? I thought this problem was based on the formula: Acceleration $\displaystyle = \frac {(v2 - v1)}{t}$, but I can't get that answer. What am I doing wrong? Or is the correct answer not listed? Thanks for any help.

Problem: A Pinewood Derby car rolls down a track. The total distance travel is measured each second. Use the chart to calculate the acceleration of the car.

Time(s) | Dist(m)

0 | 0

1 | 1

2 | 3

3 | 6

4 | 10

A) 1.0 meters/sec/sec
B) 1.5 meters/sec/sec
C) 2.0 meters/sec/sec
D) 2.5 meters/sec/sec

Your equation acceleration = (v2 - v1)/t is correct (if the acceleration is constant), but you don't have measurements of velocity. You should also have one that says distance=acceleration*t^2/2 (also for constant acceleration and a standing start). You do have measures of distance, and any of the measurements of distance will work. You also have to note that the measurements you have are distance in the last second. If you want to use the data from 4 seconds, you need the total distance traveled.

• ok, i think i got it now. The ending velocity is actually the dist traveled from sec 3 to sec 4 which is 4 meters. so V2 is 4m/s and Acc = (4m/s - 0m/s)/4 = 1 m/s/s !!! I get it now. I was calculating ave velocity for v2 as 10/4 which was wrong. – Rick S May 13 '11 at 13:54
• @Rick S: No, because of the acceleration the velocity is not constant. So you cannot just take the difference in positions from 3 sec to 4 sec to get the velocity. That gives the average velocity over that period. See my comment to Sebastien's answer about what the data means. In my way of reading the data, you should get 2 m/sec^2 for the acceleration, so since he gets the right answer it seems he is reading it correctly. – Ross Millikan May 13 '11 at 14:00

If $x(t)$ denotes the total distance travelled at time $t$, then you can check (according to your data) that

$x(t) = \dfrac{t^2+t}2$

where $x$ is in m, $t$ in s. The acceleration is given by $a = \ddot x$, hence $a = 1\,\text m/\text s/\text s$.

• I believe the data is intended to say that the car traveled 10 meters from 3.0 seconds to 4.0 seconds, as usually these cars start at rest. This would change the answer. If the data says that the car has traveled a total of 10 meters at the end of 4 seconds your solution is correct, but the car started at 0.5 m/sec. – Ross Millikan May 13 '11 at 13:54
• Thanks for your help. This was really meant as a very basic level question for kids in 5th grade or lower. The algebra is a little other their heads which is why I assumed there was an "easy" answer. – Rick S May 13 '11 at 14:01
• Ok Sebastien, I'm not sure I follow your output. I wrote question just as it was given ( kind of ambiguous ). I think Pinewood Derby cars start at rest. Forgive my lack of Algebra knowledge, but what is X with 2 dots over it? – Rick S May 13 '11 at 14:10
• $\ddot x$ is the second order derivative of the position with respect to time, which is the definition of the acceleration. However, as stated by previous remarks the calculation is correct assuming the second column of your table is the distance travelled from time 0. This means indeed that at time $t=0$, the vehicle has an initial velocity. – Sebastien May 13 '11 at 14:17
• Hey Sebastien, I totally get that your answer is correct. I just can't explain a second order derivative to a 5th grader ( or myself for that matter :) ) – Rick S May 13 '11 at 14:21

Remark: Since the data is discrete, we are going to assume mean velocities in each time intervals, otherwise the movement's law would have to be given explicitly.

Another approach is to compute the velocity first and then the acceleration, without using equations explicitly. Since you are given five pairs $(t,d)$, where $t$ is the time and $d$ is the distance, we can evaluate the average velocity $v=\frac{\Delta d}{\Delta t}$ over each period $\left[t_{i},t_{i+1}\right]$, starting at $t=0$ s, and $\Delta t_{i}=d_{i+1}-d_{i}$. The measurements are equally spaced, with $\Delta t=1$ s. We get the following average velocities:

• $i=0\quad t\in \left[ 0,1\right]$ s,$\quad v=\frac{1-0}{1}=1$ ms$^{-1}=1 \text{m/s}$,
• $i=1\quad t\in \left[ 1,2\right]$ s,$\quad v=\frac{3-1}{1}=2$ ms$^{-1}$,
• $i=2\quad t\in \left[ 2,3\right]$ s,$\quad v=\frac{6-3}{1}=3$ ms$^{-1}$,
• $i=3\quad t\in \left[ 3,4\right]$ s,$\quad v=\frac{10-6}{1}=4$ ms$^{-1}$.

Now we compute the average acceleration from one period to the next $a=\dfrac{\Delta v}{% \Delta t}$:

• from $\left[ 0,1\right]$ to $\left[ 1,2\right]$ s,$\quad a=% \frac{2-1}{1}=1$ ms$^{-2},$
• from $\left[ 1,2\right]$ to $\left[ 2,3\right]$ s,$\quad a=% \frac{3-2}{1}=1$ ms$^{-2},$
• from $\left[ 2,3\right]$ to $\left[ 3,4\right]$ s,$\quad a=% \frac{4-3}{1}=1$ ms$^{-2}.$

Thus the average acceleration $a$ is constant: $a=1$ $\text{ms}^{-2}=1\text{m/s}^{2}$.

In summary we get the following table with data and computations:

$$\begin{array}{ccccccccc} i & t_{i}\text{ (s)} & & d_{i}\text{ (m)} & & v_{i}\text{ (ms}^{-1}% \text{)} & & a_{i}\text{ (ms}^{-2}\text{)} & \\ & & & & & & & & \\ & 0 & & 0 & & & & & \\ 0 & 1 & & 1 & & 1 & & & \\ 1 & 2 & & 3 & & 2 & & 1 & \\ 2 & 3 & & 6 & & 3 & & 1 & \\ 3 & 4 & & 10 & & 4 & & 1 & \end{array}.$$

Notes:

1. Since the measurements are discrete, the velocity and accelerate values are mean values and not instant one.
2. The acceleration unit ms$^{-2}$ (or $\text{m}/\text{s}^2$) is the same as meters/sec/sec in the question, tought this is an abuse of notation, since the meaning is (meters/second)/second.
• Thanks Americo. I think this is what they were after from this question. – Rick S May 13 '11 at 15:43
• @Rick S: You are welcome! – Américo Tavares May 13 '11 at 15:45

OK, I actually think that Ross's answer makes more sense. His assumption is that the last column of your table is the distance travelled between the last two measurements. Also, he assumed (as I did) constant velocity $a$, in which case the distance travelled at time $t$ is

$x = \dfrac{at^2}2$

Now, the last column actually corresponds to

$\Delta x(t) = x(t)-x(t-\Delta t) = a\Delta t(t-\dfrac{\Delta t}2)$,

with $\Delta t$ (time between two successive measurements) is constantly $1\,\text s$, and $\Delta x$ is the distance travelled in this interval of time (in other words: the second column of the table). Then I think you do find that $a = 2\,\text m.\text s^{-2}$.