Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 212: 28

$\approx 10\;\rm m/s^2$

First, we need to plot these data into a graph, as shown in the first figure below. It is obvious that the plane is moving toward the east starting from 90 m away from our origin and at the same time, it is moving toward the north. Thus, it is moving northwest from the start point. To find the plane's acceleration we need to find its $a_x$ and $a_y$ accelerations. Let's assume that the plane is moving at a constant acceleration, so that $$x=x_0+v_{0x}t+\frac{1}{2}a_xt^2$$ we know that the plane accelerates from rest, so $$x=90+(0)t+\frac{1}{2}a_xt^2$$ $$x=90+ \frac{1}{2}a_xt^2$$ we will use $t^2$ as a function of $x$ where the model of $y=a+bx$ is used at which $b$ is the slope of the straight line. Thus, drawing a graph of $x$ versus $t^2$ as a straight line at which its slope is given by $${\rm slope_1}=\frac{1}{2}a_x\tag 1$$ \begin{array}{|c|c|c|} \hline t^2& x \\ \hline 0^2=0 & 91 \\ \hline 1^2=1 & 86 \\ \hline 2^2=4 & 77 \\ \hline 3^2=9 & 65 \\ \hline 4^2=16&39 \\ \hline 5^2=25& 19 \\ \hline \end{array} The slope of the second figure below, which is given by the best-fit line as we see below, is $${\rm slope_1}=\frac{1}{2}a_x $$ Thus, $$a_x=2\;{\rm slope_1}=2\times -2.94 =\color{blue}{\bf -5.88}\;\rm m/s^2$$ By the same approach; $${\rm slope_2}=\frac{1}{2}a_y\tag 2$$ \begin{array}{|c|c|c|} \hline t^2& x \\ \hline 0^2=0 & 0\\ \hline 1^2=1 & 4\\ \hline 2^2=4 & 18\\ \hline 3^2=9 & 39\\ \hline 4^2=16&63\\ \hline 5^2=25& 101\\ \hline \end{array} The slope of the third figure below, which is given by the best-fit line as we see below, is $${\rm slope_2}=\frac{1}{2}a_y $$ Thus, $$a_y=2\;{\rm slope_2}=2\times 4.02=\color{blue}{\bf 8.04} \;\rm m/s^2$$ Therefore, $$a=\sqrt{a_x^2+a_y^2}=\sqrt{(-5.88)^2+8.04^2}=\color{red}{\bf 9.96}\;\rm m/s^2$$