Answer: The speed of the puck after it leaves the stick is 45.94 meter per second. The distance traveled by the puck while it is in contact with the stick is 0.51 meter per second.
Explanation:
Convert time from ms to s.
[tex]1 ms =10^{-3} s[/tex]
[tex]t=22.4\times 10^{-3} s[/tex]
Calculate the force by putting the values of a,b and c in the expression of given expression of the force.
[tex]F(t)=a+bt+ct^{2}[/tex]
Put a=−25.0 N, [tex]b=1.11\times 10^{5} Ns^{-1}[/tex],[tex]t=22.4\times 10^{-3} s[/tex] and [tex]c=-5.58\times 10^{6} Ns^{-2}[/tex].
[tex]F(t)=−25.0+(1.11\times 10^{5})(22.4\times 10^{-3})+(-5.58\times 10^{6})(22.4\times 10^{-3})^{2}[/tex]
[tex]F(t)=-338.42 N[/tex]
Use the formula of force in terms of acceleration.
[tex]F=ma[/tex]
Put m=165 g.
[tex]-338.42=(165 g)(\frac{1 kg}{1000 g})a[/tex]
[tex]a=-2051.03 ms^{-2}[/tex]
The negative sign in the acceleration represents the deceleration.
Use the expression of the acceleration to calculate the speed of the puck after it leaves the stick.
[tex]a=\frac{v-u}{t}[/tex]
Here, v is the final speed and u is the initial speed.
Put u=0, a=-2051.03 meter per second and [tex]t=22.4\times 10^{-3} s[/tex]
.
[tex]-2051.03=\frac{v-0}{22.4\times 10^{-3}}[/tex]
[tex]v=-45.94 ms^{-1}[/tex]
To determine how far the puck travels while it is not in contact with stick by using equation of motion.
[tex]S=ut^{2}+\frac{1}{2}at^{2}[/tex]
Here, S is the distance traveled by the object, u is the initial speed and t is the time.
Put [tex]v=-45.94 ms^{-1}[/tex], [tex]t=22.4\times 10^{-3} s[/tex] and a=2051.03 meter per second.
[tex]S=(0)(22.4\times 10^{-3})+\frac{1}{2}(2051.03)(22.4\times 10^{-3})^{2}[/tex]
[tex]S=frac{1}{2}(2051.03)(22.4\times 10^{-3})^{2}[/tex]
[tex]S=0.51 m[/tex]
The speed of the puck after it leaves the stick is 45.94 meter per second. The distance traveled by the puck while it is in contact with the stick is 0.51 meter per second.
