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I am a student in an Introduction to Physics class, and I have the following question which I am confused about, and how two parts of the answers relate to each other. (We are doing 1d kinematics)

A car traveling 45 $\frac{km}{h}$ slows down at a constant 0.5 $\frac{m}{s^2}$ just by "letting up on the gas." Calculate the distance it travels during the first and fifth second.

The source, I believe, is from the fifth edition of the Giancoli Physics book, and I have only included part (c) because that is the part relevant to the question.

I did some work and using $x=v_ot+\frac{1}{2}at^2$, I did $x_1-x_0$ to obtain the distance traveled in the first second (12.25 m), which also is just $x_1$. I did the same thing with $x_5-x_4$ to obtain the distance traveled in the fifth second (10.25 m). Now, after I did this, I noticed a small thing.

The difference traveled during the first second and the fifth second is $-2$ meters with a time span of 4 seconds, and the acceleration is -0.5 $\frac{m}{s^2}$. I related $-2=4 \cdot -0.5$.

I am wondering why this is true. I am also confused as to why acceleration plays a part in this question: doesn't the distance traveled during constant acceleration grow and decrease like a parabola? Why is it linear in this case?

Thank you so much!

EDIT: I do not believe that this is a coincidence (at least I hope not) because I tried this with $x_4-x_3$, but please correct me if I am wrong.

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  • $\begingroup$ Were you to make a plot of this function from 0 to 5 seconds, you would find that it would appear very flat, but it is still curved like a parabola. $\endgroup$ – Triatticus Sep 19 '20 at 23:13
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In a given time interval $T = [t_1,t_2]$, the distance travelled during that interval is given by $\Delta x_T = \int_{t_1}^{t_2} v(t) \mathrm{d}t = \bar v_T \cdot(t_2-t_1)$, where $\bar v$ is the average velocity over the time interval. If the acceleration is constant, then $\bar v_T = \frac{v(t_1) + v(t_2)}{2} = v(\frac{t_1 + t_2}{2})$. Thus, the distance travelled during an interval is equal to the velocity in the middle of that interval times the "length" of the interval.

Given two time intervals of equal size, $T = [t_1,t_2]$ and $T^\prime = [t_1+t^\prime,t_2+t^\prime]$, the difference in the distance travelled during each of them is given by $$\Delta x_{T^\prime} - \Delta x_T = (\bar v_{T^\prime} - \bar v_T) \cdot (t_2 - t_1) = \left( v(\frac{t_1+t_2}{2} + t^\prime) - v(\frac{t_1+t_2}{2}) \right) \cdot (t_2 - t_1)$$ Note that $v(\frac{t_1+t_2}{2} + t^\prime) = v(\frac{t_1+t_2}{2}) + at^\prime$, so if you were to divide this expression by $t^\prime$, the time between the intervals, you would get $$\frac{\Delta x_{T^\prime} - \Delta x_T}{t^\prime} = a\cdot(t_2-t_1)$$ The reason this relation works is because the distance travelled during an interval of fixed "length" is proportional to velocity, which scales linearly with time when acceleration is constant. Also, note that your equation only worked because your time intervals were $1 \,\rm s$ long; in general, you need to account for the size of the interval.

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I do not think that anyone will care, but I found a more elementary way, using solely manipulation and $x=v_ot+\frac{1}{2}at^2$. I followed from Sandejo's solution using $[t,t_1]$ and $[t+T,t_1+T]$.

On the interval $[t,t_1]$,

\begin{align*} x_{t} &= v_0t+\frac{1}{2}at^2 \\ x_{t_1} &= v_0t_1+\frac{1}{2}a(t_1)^2 \\ x_{t-1}-x_{t} = x_1 &= v_0(t_1-t)+\frac{1}{2}a(t_1-t)(t_1+t) \\ x_1 &= (t_1-t)(v_0+\frac{1}{2}a(t_1+t)) \\ \end{align*}

On the interval $[t+T, t_1+T]$,

\begin{align*} x_{t+T} &= v_0(t+T)+\frac{1}{2}a(t+T)^2 \\ x_{t_1+T} &= v_0(t_1+T)+\frac{1}{2}a(t_1+T)^2 \\ x_{t_1+T}-x_{t+T}= x_2 &= v_0(t_1-t)+\frac{1}{2}a(t_1-t)(t_1+t+2T) \\ x_2 &= (t_1-t)(v_0+\frac{1}{2}a(t_1+t+2T)) \\ \end{align*}

Then,

\begin{align*} \frac{x_1}{t_1-t} &= v_0+\frac{1}{2}a(t_1+t) \\ \frac{x_2}{t_1-t} &= v_0+\frac{1}{2}a(t_1+t+2T) \\ \end{align*}

Finally, by dividing, $$\frac{x_2-x_1}{t_1-t}=aT$$

Since $t_1-t=1$ for this problem, then $x_2-x_1=aT$, and this is linear. I hope that this makes sense.

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