Answer:
a) [tex]a_{sled} = 0.619\,\frac{m}{s^{2}}[/tex], b) [tex]a_{girl} = 0.13\,\frac{m}{s^{2}}[/tex], c) [tex]x_{girl} = 2.604\,m[/tex]
Explanation:
a) The acceleration magnitude for the sled is:
[tex]a_{sled} = \frac{5.2\,N}{8.4\,kg }[/tex]
[tex]a_{sled} = 0.619\,\frac{m}{s^{2}}[/tex]
b) The acceleration magnitude for the girl is:
[tex]a_{girl} = \frac{5.2\,N}{40\,kg }[/tex]
[tex]a_{girl} = 0.13\,\frac{m}{s^{2}}[/tex]
c) The position equations for the girl and sled are, respectively:
[tex]x_{girl} = 0\,m + \frac{1}{2}\cdot (0.13\,\frac{m}{s^{2}} )\cdot t^{2}[/tex]
[tex]x_{sled} = 15\,m-\frac{1}{2}\cdot (0.619\,\frac{m}{s^{2}} )\cdot t^{2}[/tex]
It is needed to know in which time the girl and sled meet each other. That is:
[tex]0\,m + \frac{1}{2}\cdot (0.13\,\frac{m}{s^{2}} )\cdot t^{2} =15\,m-\frac{1}{2}\cdot (0.619\,\frac{m}{s^{2}} )\cdot t^{2}[/tex]
[tex]15\,m - \frac{1}{2}\cdot (0.749\,\frac{m}{s^{2}} )\cdot t^{2} = 0[/tex]
[tex]t \approx 6.329\,s[/tex]
The final position in comparison with the initial position of the girl is:
[tex]x_{girl} = 0\,m + \frac{1}{2}\cdot (0.13\,\frac{m}{s^{2}} )\cdot (6.329\,s)^{2}[/tex]
[tex]x_{girl} = 2.604\,m[/tex]
