Solved Problem on One-dimensional Motion
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A car travels with constant acceleration, initial speed v0, and acceleration α.
a) Calculate the distance traveled by car at n-th second (that is, between the instants n-1 and n);
b) For v0 = 15 m/s and α = 1.2 m/s2, calculate the distance traveled in the first second and in the fifteenth second.


Problem data:
  • Initial speed:    v0;
  • Acceleration:    α.
Problem diagram:

We choose a reference frame pointing to the right, Sn is the displacement of the car from the origin to the n-th second, and Sn−1 is the displacement from the origin to the previous second (Figure 1).

Figure 1

Solution

a) The equation for displacement as a function of time with constant acceleration is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {S=S_{0}+v_{0}t+\frac{a}{2}t^{2}} \tag{I} \end{gather} \]
Applying the expression (I) to the car displacement up to instant t = n−1
\[ S_{n-1}=S_{0}+v_{0}(n-1)+\frac{\alpha }{2}(n-1)^{2} \]
From Special Binomial Products
\[ (a-b)^{2}=a^{2}-2ab+b^{2} \]

applying to the term \( (n-1)^{2} \)
\[ \begin{gather} S_{n-1}=S_{0}+v_{0}(n-1)+\frac{\alpha}{2}(n-1)^{2}\\[5pt] S_{n-1}=S_{0}+v_{0}n-v_{0}+\frac{\alpha}{2}(n^{2}-2n+1)\\[5pt] S_{n-1}=S_{0}+v_{0}n-v_{0}+\frac{\alpha}{2}n^{2}-\frac{\alpha }{2}2n+\frac{\alpha}{2}\\[5pt] S_{n-1}=S_{0}+v_{0}n-v_{0}+\frac{\alpha }{2}n^{2}-\alpha n+\frac{\alpha }{2} \tag{II} \end{gather} \]
Applying the expression (I) for instant t = n
\[ \begin{gather} S_{n}=S_{0}+v_{0}n+\frac{\alpha }{2}n^{2} \tag{III} \end{gather} \]
We want the displacement only at the n-th second, we have the condition.
\[ \Delta S=S_{n}-S_{n-1} \]
subtracting the expression (II) from (III)
\[ \begin{gather} \Delta S=S_{0}+v_{0}n+\frac{\alpha}{2}n^{2}-\left[S_{0}+v_{0}n-v_{0}+\frac{\alpha }{2}n^{2}-\alpha n+\frac{\alpha }{2}\right]\\ \Delta S=S_{0}+v_{0}n+\frac{\alpha}{2}n^{2}-S_{0}-v_{0}n+v_{0}-\frac{\alpha }{2}n^{2}+\alpha n-\frac{\alpha }{2}\\ \Delta S=v_{0}+\alpha n-\frac{\alpha}{2} \end{gather} \]
factoring α
\[ \bbox[#FFCCCC,10px] {\Delta S=v_{0}+\alpha \left(n-\frac{1}{2}\right)} \]

b) Using the expression obtained in the previous item and the data values
  • For n = 1
\[ \begin{gather} \Delta S=15+1.2\times \left(1-\frac{1}{2}\right)\\ \Delta S=15+1.2\times \left(\frac{2-1}{2}\right)\\ \Delta S=15+1.2\times \frac{1}{2}\\ \Delta S=15+0.6 \end{gather} \]
\[ \bbox[#FFCCCC,10px] {\Delta S=15.6\;\text{m}} \]

Figure 2
  • For n = 15
\[ \begin{gather} \Delta S=15+1.2\times \left(15-\frac{1}{2}\right)\\ \Delta S=15+1.2\times \left(\frac{30-1}{2}\right)\\ \Delta S=15+1.2\times \frac{29}{2}\\ \Delta S=15+0.6\times 29\\ \Delta S=15+17.4 \end{gather} \]
\[ \bbox[#FFCCCC,10px] {\Delta S=32.4\;\text{m}} \]

Figure 3

Note: In the first case, the car travels 15.6 m, this is also the displacement from the initial instant to t = 1 s. In the second case, the car travels 32.4 m between t = 14 s and t = 15 s, but this is not the point of the trajectory in which he has been since he left the origin.
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .