Solved Problem on Two-dimensional Motion and Relative Motion

Solved Problem on Two-dimensional Motion

From the vertex of a right angle, two drivers start, with a time interval equal to n seconds, moving at constant speeds along the two sides. Calculate the speeds of the two drivers, knowing that after t seconds from the departure of the second driver, their distance is d, and after T seconds, it is D.

Problem data:

Time interval between the departures of the two drivers: n;
Distance between drivers after t seconds: d;
Distance between drivers after T seconds: D.

Problem diagram:

We choose a reference frame with 2 perpendicular axes. The first driver starts from the origin with speed v₁ in the x direction, and after n seconds the second driver starts from the origin with constant speed v₂ in the y direction. During the interval of n seconds, the first driver will have covered a distance equal to v₁.n; this distance will be the initial position of the first driver when the time count starts, and the second driver starting from the origin will have an initial position equal to zero (Figure 1).

Figure 1

Solution

The drivers have constant speeds, and they describe a Uniform Rectilinear Motion, the equation for this motion is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {S=S_0+v\tau} \tag{I} \end{gather} \]

Note: Here the time is represented by τ instead of t, generally used, to avoid confusion with the time interval t given in the problem.

Writing the equation (I) for drivers 1 and 2 in the time intervals t and T

\[ \begin{gather} S_1(\tau)=S_{01}+v_1\tau\\[5pt] S_1(t)=v_1 n+v_1 t=v_1(n+t) \tag{II}\\[5pt] S_1(T)=v_1 n+v_1 T=v_1(n+T) \tag{III} \end{gather} \]

\[ \begin{gather} S_2(\tau)=S_{02}+v_2\tau\\[5pt] S_2(t)=0+v_2t=v_2t \tag{IV}\\[5pt] S_2(T)=0+v_2T=v_2T \tag{V} \end{gather} \]

In Figure 2, we have S₁(t) as the distance traveled by driver 1 in the time interval t, and S₂(t) as the distance traveled by driver 2 in this time interval, using the Pythagorean Theorem

\[ \begin{gather} h^2=S_1(t)^2+S_2(t)^2 \tag{VI} \end{gather} \]

Substituting equations (II), (III), (IV), and (V) into condition (VI)

\[ \left\{ \begin{array}{l} d^2=[v_1(n+t)]^2+v_2^2t^2\\ D^2=[v_1(n+T)]^2+v_2^2T^2 \tag{VII} \end{array} \right. \]

this is a system of two equations with two unknowns, v₁, and v₂, solving the first equation of the system (VII) for \( v_2^2 \)

\[ \begin{gather} v_2^2=\frac{d^2-[v_1(n+t)]^2}{t^2} \tag{VIII} \end{gather} \]

substituting this value into the second equation of the system (VII)

\[ \begin{gather} D^2=[v_1(n+T)]^2+\left\{\frac{d^2-[v_1(n+t)]^2}{t^2}\right\}T^2 \end{gather} \]

Figure 2

multiplying this equation by t₂

\[ \begin{gather} D^2t^2=[v_1(n+T)]^2t^2+\left\{\frac{d^2-[v_1(n+t)]^2}{\cancel{t^2}}\right\}T^2\cancel{t^2}\\[5pt] D^2t^2=[v_1(n+T)]^2t^2+\left\{d^2-[v_1(n+t)]^2\right\}T^2\\[5pt] D^2t^2=[v_1(n+T)]^2t^2+d^2T^2-[v_1(n+t)]^2T^2\\[5pt] D^2t^2-d^2T^2=v_1^2(n+T)^2t^2-v_1^2(n+t)^2T^2 \end{gather} \]

factoring out \( v_1^2 \) on the right side of the equation

\[ \begin{gather} D^2t^2-d^2T^2=v_1^2[(n+T)^2t^2-(n+t)^2T^2] \end{gather} \]

From the Special Binomial Products

\[ \begin{gather} (a+b)^2=a^2+2ab+b^2 \end{gather} \]

applying this product to terms \( (n+T)^2 \) and \( (n+t)^2 \)

\[ \begin{gather} D^2t^2-d^2T^2=v_1^2[(n^2+2nT+T^2)t^2-(n^2+2nt+t^2)T^2]\\[5pt] D^2t^2-d^2T^2=v_1^2[n^2t^2+2nTt^2+T^2t^2-n^2T^2-2ntT^2-t^2T^2]\\[5pt] D^2t^2-d^2T^2=v_1^2[n^2t^2+2nTt^2-n^2T^2-2ntT^2] \end{gather} \]

factoring out n² and 2ntT in the term in brackets

\[ \begin{gather} D^2t^2-d^2T^2=v_1^2[n^2(t^2-T^2)+2ntT(t-T)] \end{gather} \]

From the Special Binomial Products

\[ \begin{gather} \left(a^2-b^2\right)=(a+b)(a-b) \end{gather} \]

applying this product to the term \( \left(t^2-T^2\right) \)

\[ \begin{gather} D^2t^2-d^2T^2=v_1^2[n^2(t+T)(t-T)+2ntT(t-T)] \end{gather} \]

factoring out the term n(t-T) inside the brackets

\[ \begin{gather} D^2t^2-d^2T^2=v_1^2\left\{n(t-T)\left[n(t+T)+2tT\right]\right\}\\[5pt] v_1^2=\frac{D^2t^2-d^2T^2}{n(t-T)\left[n(t+T)+2tT\right]} \tag{IX} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {v_1=\pm\sqrt{\frac{D^2t^2-d^2T^2}{n(t-T)\left[n(t+T)+2tT\right]}}} \end{gather} \]

Substituting this value of \( v_1^2 \), given in the form of equation (IX), into equation (VIII) and factoring out the term \( \frac{1}{t^2} \), we will have the value of v₂

\[ \begin{gather} v_2^2=\frac{d^2-[v_1(n+t)]^2}{t^2}\\[5pt] v_2^2=\frac{1}{t^2}\left[d^2-v_1^2(n+t)^2\right]\\[5pt] v_2^2=\frac{1}{t^2}\left\{d^2-\frac{D^2t^2-d^2T^2}{n(t-T)\left[n(t+T)+2tT\right]}(n+t)^2\right\} \end{gather} \]

the term \( n(t-T)\left[n(t+T)+2tT\right] \) is the common factor of the terms in the braces

\[ \begin{gather} v_2^2=\frac{1}{t^2}\left\{\frac{d^2\left[n(t-T)\left[n(t+T)+2tT\right]\right]-(D^2t^2-d^2T^2)(n+t)^2}{n(t-T)\left[n(t+T)+2tT\right]}\right\}\\[5pt] v_2^2=\frac{1}{t^2}\left\{\frac{d^2\left[n^2(t+T)(t-T)+2tTn(t-T)\right]-(D^2t^2-d^2T^2)(n+t)^2}{n(t-T)\left[n(t+T)+2tT\right]}\right\}\\[5pt] v_2^2=\frac{1}{t^2}\left\{\frac{d^2\left[n^2(t^2-T^2)+2t^2Tn-2tT^2n\right]-(D^2t^2-d^2T^2)(n+t)^2}{n(t-T)\left[n(t+T)+2tT\right]}\right\}\\[5pt] v_2^2=\frac{1}{t^2}\left\{\frac{d^2\left[n2t^2-n^2T^2+2t^2Tn-2tT^2n\right]-(D^2t^2-d^2T^2)(n+t)^2}{n(t-T)\left[n(t+T)+2tT\right]}\right\} \end{gather} \]

in the term in brackets \( n^2t^2-n^2T^2+2t^2Tn-2tT^2n \), we will group the terms as follows, \( (n^2t^2+2t^2Tn)+(-n^2T^2-2tT^2n). \)

\[ (n^2t^2+2t^2Tn)+(-n^2T^2-2tT^2n) \]

. In the first term in parentheses, we factor out t₂, and in the second term, \( -T^2 \)

\[ \begin{gather} v_2^2=\frac{1}{t^2}\left\{\frac{d^2\left[(n^2t^2+2t^2Tn)+(-n^2T^2-2tT^2n)\right]-(D^2t^2-d^2T^2)(n+t)^2}{n(t-T)\left[n(t+T)+2tT\right]}\right\}\\[5pt] v_2^2=\frac{1}{t^2}\left\{\frac{d^2\left[t^2(n^2+2Tn)-T^2(n^2+2tn)\right]-(D^2t^2-d^2T^2)(n+t)^2}{n(t-T)\left[n(t+T)+2tT\right]}\right\} \end{gather} \]

in the term in brackets, we will add and subtract T₂ in the first term in parentheses, and in the second term in parentheses, we will add and subtract t₂.

Note: The terms in parentheses \( n^2+2Tn \) and \( n^2+2tn \) are similar to the Special Binomial Product \( a^2+2ab+b^2=(a+b)^2 \), except for the missing term b² (T² in the first equation and t² in the second equation).
Then we can use the Completing the Square Method, if we have an equation of the form \( a^2+2ab \), adding and subtracting the square of the missing term does not change anything, we are adding zero to the equation. \( a^2+2ab+\underbrace{b^2-b^2}_{0} \), The first three terms form a Special Binomial Product \( \underbrace{a^2+2ab+b^2}_{(a+b)^2}-b^2 \), so that we can write our original equation in the form \( a^2+2ab=(a+b)^2-b^2 \).

\[ \begin{gather} v_2^2=\frac{1}{t^2}\left\{\frac{d^2\left[(n^2+2Tn+T^2-T^2)t^2-(n^2+2tn+t^2-t^2)T^2\right]-(D^2t^2-d^2T^2)(n+t)^2}{n(t-T)\left[n(t+T)+2tT\right]}\right\} \end{gather} \]

the terms \( n^2+2Tn+T^2 \) and \( n^2+2tn+t^2 \) are in the same form as the Special Binomial Product \( (a+b)^2 \)

\[ \begin{gather} v_2^2=\frac{1}{t^2}\left\{\frac{d^2\left[((n+T)^2-T^2)t^2-((n+t)^2-t^2)T^2\right]-(D^2t^2-d^2T^2)(n+t)^2}{n(t-T)\left[n(t+T)+2tT\right]}\right\}\\[5pt] v_2^2=\frac{1}{t^2}\left\{\frac{d^2\left[(n+T)^2t^2-T^2t^2-(n+t)^2T^2+t^2T^2\right]-(D^2t^2-d^2T^2)(n+t)^2}{n(t-T)\left[n(t+T)+2tT\right]}\right\}\\[5pt] v_2^2=\frac{1}{t^2}\left\{\frac{d^2\left[(n+T)^2t^2-(n+t)^2T^2\right]-(D^2t^2-d^2T^2)(n+t)^2}{n(t-T)\left[\;n(t+T)+2tT\right]}\right\}\\[5pt] v_2^2=\frac{1}{t^2}\left\{\frac{d^2(n+T)^2t^2-d^2(n+t)^2T^2-D^2(n+t)^2t^2+d^2(n+t)^2T^2}{n(t-T)\left[n(t+T)+2tT\right]}\right\}\\[5pt] v_2^2=\frac{1}{t^2}\left\{\frac{d^2(n+T)^2t^2-D^2(n+t)^2t^2}{n(t-T)\left[n(t+T)+2tT\right]}\right\}\\[5pt] v_2^2=\frac{d^2(n+T)^2-D^2(n+t)^2}{n(t-T)\left[n(t+T)+2tT\right]} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {v_2=\pm\sqrt{\frac{d^2(n+T)^2-D^2(n+t)^2}{n(t-T)\left[n(t+T)+2tT\right]}}} \end{gather} \]

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