Answer:
a) [tex]v(7s)=46.38m/s[/tex]
b) [tex]t=70s[/tex]
c) [tex]v(70s)=-58.2m/s[/tex]
Explanation:
From the exercise we know the arrow's equation of position
[tex]s(t)=58t-0.83t^2[/tex]
a) If we want to know the velocity of the arrow after t=7s we need to derivate the equation of position to get velocity
[tex]v(t)=\frac{ds}{dt}=58-1.66t[/tex]
Now, we evaluate 7s in the equation
[tex]v(7s)=58m/s-(1.66m/s^2)(7s)=46.38m/s[/tex]
b) To find how much time does it take the arrow to hit the moon we need to evaluate s=0 at the equation of position in the vertical direction
[tex]s(t)=58t-0.83t^2[/tex]
[tex]0=58t-0.83t^2[/tex]
Solving the quadratic equation using the following formula
[tex]t=\frac{-b±\sqrt{b^2-4ac}}{2a}[/tex]
[tex]a=-0.83\\b=58\\c=0[/tex]
[tex]t=0[/tex] or [tex]t=70s[/tex]
Since time can not be zero the answer is t=70s
c) To calculate the arrow's velocity when it hits the moon we need to evaluate t=70s in the equation of velocity
[tex]v(70s)=58m/s-(1.66m/s^2)(70s)=-58.2m/s[/tex]
