Answer:
Explanation:
We can use the following kinematics equation
[tex]v_f^2-v_i^2=2 \ a \ d[/tex]
where [tex]v_f[/tex] is the final speed, [tex]v_i[/tex] its the initial speed, a is the acceleration, and d the distance.
The force will be tripled, the force is:
[tex]\vec{F} = m \vec{a}[/tex]
in 1D
[tex]F = m a[/tex]
Now, for the original problem, we have
[tex](25 \frac{m}{s})^2-(0 \frac{m}{s})^2=2 \ a' \ d'[/tex]
[tex](25 \frac{m}{s}))^2=2 \ a' \ d'[/tex]
[tex]625 \frac{m^2}{s^2}=2 \ a' \ d'[/tex]
For the second problem, we have
[tex](v_f})^2-(v_i)^2=2 \ a'' \ d''[/tex]
starting from the rest, we have the same initial velocity.
[tex](v_f})^2-(0\frac{m}{s})^2=2 \ a'' \ d''[/tex]
[tex](v_f})^2=2 \ a'' \ d''[/tex]
As the force is tripled, we have:
[tex]F'' = 3 F'[/tex]
[tex]m'' \ a'' = 3 m' \ a'[/tex]
But the mass its the same, so
[tex]m' \ a'' = 3 m' \ a'[/tex]
[tex] a'' = 3 \ a'[/tex]
So the acceleration its also tripled.
[tex](v_f})^2=2 \ (3 * a') \ d''[/tex]
[tex](v_f})^2=3 ( 2 \ ( a' \ d'' )[/tex]
As the distance traveled by the arrow must also be the same, we have:
[tex](v_f})^2=3 ( 2 \ ( a' \ d' )[/tex]
[tex](v_f})^2=3 (625 \frac{m^2}{s^2})[/tex]
[tex]v_f= \ sqrt{3 (625 \frac{m^2}{s^2})}[/tex]
[tex]v_f= \ sqrt{3} 25 \frac{m}{s}[/tex]
[tex]v_f= 43.30 \frac{m}{s}[/tex]
And this will be the speed from the arrow leaving the bow.
