## Arrow's initial velocity (Force problem)

1. The problem statement, all variables and given/known data

An arrow, starting from rest, leaves the bow with a speed of 27.5 m/s. If the average force exerted on the arrow by the bow was increased 3 times and the arrow was accelerated over the same distance, then with what speed would the arrow leave the bow?

2. Relevant equations

F = ma, obviously. So if 3F = 3ma, then the acceleration would triple, because the mass of the arrow remains the same.

Vx=V0+at

3. The attempt at a solution

3*27.5 is wrong, so tripling the force doesn't automatically triple the initial velocity. I don't feel like I have enough information to work this problem, honestly.

Does anyone have any suggestions or help to offer?
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 Recognitions: Homework Help You're right that acceleration triples. But you don't know the time... you know the distance is the same in both situations. There's another kinematics equation that's useful in this circumstance. hint: it doesn't have time in it.
 So using v^2 = v0^2 + 2ax, should I find the final velocity the first time? Using that the acceleration triples could give: v^2 = v0^2 + 6ax And the first launch is v^2 = 27.5^2 + 2ax. The distance is the same, so if one solves for x and sets the equations equal you get: (v^2-v0^2)/6a =( v^2-(27.5^2))/2a But that's still too many variables to solve! How am I supposed to know the final velocity and/or acceleration?

Recognitions:
Homework Help

## Arrow's initial velocity (Force problem)

In the first situation 27.5 is the final velocity. v0 = 0.
In the second situation, you're trying to calculate final velocity, and v0 = 0.

Write the two kinematics equations for the two situations... then try to see if you can eliminate a and x.

we're looking over the period in which the bow is accelerated... so the velocity with which the bow leaves, is the final velocity (at the end of the acceleration)
 Well x is easily eliminated since it is the same in both equations. Solving for x and then setting the two equal to each other gives: 27.5^2/(2a) = v^2/(6a) Rearranging and solving for v^2 gives: 2268.75a = v^2 But I'm still stuck with a, and have no idea where it could go or how to get rid of it.

Recognitions:
Homework Help
 Quote by thatgirlyouknow Well x is easily eliminated since it is the same in both equations. Solving for x and then setting the two equal to each other gives: 27.5^2/(2a) = v^2/(6a)
Yes, exactly.

 Rearranging and solving for v^2 gives: 2268.75a = v^2
No, you made a mistake in your algebra. You're almost there...
 Oops, the a's cancel. My mistake. So v = 47.63 m/s. Thank you so much! Problem solved. :D

Recognitions:
Homework Help
 Quote by thatgirlyouknow Oops, the a's cancel. My mistake. So v = 47.63 m/s. Thank you so much! Problem solved. :D
yup. that's the answer.

no prob. you're welcome.
 I have different question; What distance an arrow travel if I throw the arrow with 200N force and 20 degrees? The mass of the arrow is 100 gr. Please help,Thanks