Solved Problem on One-dimensional Motion

The equation of motion is

S = 21 − 10 t + t²

where position (S) is measured in meters and time (t) is measured in seconds:
a) Make a table with values for t from 0 to 8 s, and from the table plot the graph of the function;
b) When does the particle pass through the origin of the coordinate system?
c) When does the particle change its direction?

Solution

a) The table for the given function will be

t (s)	\( S=21-10t+t^2 \)	\( S(t) \)
0	\( S(0)=21-10\times 0+0^2 \)	21
1	\( S(1)=21-10\times 1+1^2 \)	12
2	\( S(2)=21-10\times 2+2^2 \)	5
3	\( S(3)=21-10\times 3+3^2 \)	0
4	\( S(4)=21-10\times 4+4^2 \)	− 3
5	\( S(5)=21-10\times 5+5^2 \)	− 4
6	\( S(6)=21-10\times 6+6^2 \)	− 3
7	\( S(7)=21-10\times 7+7^2 \)	0
8	\( S(8)=21-10\times 8+8^2 \)	5

plotting the points on the graph (Graph 1)

Graph 1

b) The particle goes through the origin of the coordinate system when S=0, from the table and graph we see that this occurs at the instants t = 3 s and t = 7 s ;

c) The particle changes direction when the sign of its speed changes. In a graph of position-time, S = f(t), the velocity is given by the tangent to the graph. From Graph 2, we see that for the points to the left of the vertex, V, the tangent is negative (tan θ < 0), so its velocity is less than zero (v < 0). For the points to the right of the parabola the tangent, has a positive sign (tan θ > 0), so its velocity is greater than zero (v>0).

The x coordinate of a vertex of a parabola is given by

\[ \begin{gather} x=-\frac{b}{2a} \end{gather} \]

in the function given in the problem, we have a=1 and b=−10, the instant the particle changes direction is

\[ \begin{gather} t=-\frac{-10}{2\times 1} \end {gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {t=5\;\mathrm s} \end{gather} \]

Graph 2

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .