Answer:
Raul will be 30 meters to the right of observer.
Explanation:
Raul is walking towards the observer, let be [tex]x[/tex] the position of Raul with respect to the observer and, besides, that [tex]x[/tex] is negative to the left and positive to the right.
Raul is moving at constant rate so that velocity ([tex]v[/tex]) is equal to:
[tex]v = \frac{x_{f}-x_{o}}{\Delta t}[/tex]
Where:
[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final position of Raul regarding the observer, measured in meters.
[tex]\Delta t[/tex] - Change in time, measured in seconds.
Given that [tex]x_{o} = -15\,m[/tex], [tex]x_{f} = -6\,m[/tex] and [tex]\Delta t = 3\,s[/tex], the velocity of Raul is:
[tex]v = \frac{(-6\,m)-(-15\,m)}{3\,s}[/tex]
[tex]v = +3\,\frac{m}{s}[/tex]
The equation of his displacement is represented by clearing final position in the equation above:
[tex]x_{f} = x_{o}+v\cdot \Delta t[/tex]
If [tex]x_{o} = -15\,m[/tex] and [tex]v = +\frac{3}{5}\,\frac{m}{s}[/tex], the equation of motion is:
[tex]x_{f} = -15\,m + \left(+3\,\frac{m}{s} \right)\cdot \Delta t[/tex]
Then, if [tex]\Delta t = 15\,s[/tex], the final position of Raul regarding the observer is:
[tex]x_{f} = -15\,m + \left(+3\,\frac{m}{s} \right)\cdot (15\,s)[/tex]
[tex]x_{f} = +30\,m[/tex]
Raul will be 30 meters to the right of observer.
