Answer :
Answer:
a) [tex]x_{f} = 3.126\,m[/tex], b) [tex]v_{f} = 3.126\,\frac{m}{s}[/tex], c) [tex]x_{f} = 21.882\,m[/tex], d) [tex]v_{f} = 6.252\,\frac{m}{s}[/tex]
Explanation:
a) Position of the puck:
The acceleration experimented by the puck is:
[tex]a_{puck} = \frac{0.25\,N}{0.160\,\frac{m}{s^{2}} }[/tex]
[tex]a_{puck} = 1.563\,\frac{m}{s^{2}}[/tex]
As the force remains constant during its time of application, accelaration is also constant. Position at given time is the following:
[tex]x_{f} = \frac{1}{2}\cdot (1.563\,\frac{m}{s^{2}} )\cdot (2\,s)^{2}[/tex]
[tex]x_{f} = 3.126\,m[/tex]
b) Speed of the puck:
The speed of the puck is computed as follows:
[tex]v_{f} = (1.563\,\frac{m}{s^{2}} )\cdot (2\,s)[/tex]
[tex]v_{f} = 3.126\,\frac{m}{s}[/tex]
c) The absence of external force and the fact that ground is frictionless lead to the conclusion that hockey puck moves out at constant speed from 2 s. to 5 s. Then, the initial speed is:
[tex]x_{o} = 3.126\,m + (3.126\,\frac{m}{s} )\cdot (5\,s-2\,s)[/tex]
[tex]x_{o} = 12.504\,m[/tex]
Likewise, the initial speed is [tex]3.126\,\frac{m}{s}[/tex].
The new application of the same force means the return of a accelerated movement. Then:
[tex]x_{f} = 12.504\,m+(3.126\,\frac{m}{s} )\cdot (7\,s-5\,s)+\frac{1}{2}\cdot (1.563\,\frac{m}{s^{2}} )\cdot (7\,s-5\,s)^{2}[/tex]
[tex]x_{f} = 21.882\,m[/tex]
d) The final speed of the puck is:
[tex]v_{f} = 3.126\,\frac{m}{s} + (1.563\,\frac{m}{s^{2}})\cdot (7\,s-5\,s)[/tex]
[tex]v_{f} = 6.252\,\frac{m}{s}[/tex]