Finding acceleration and time in force problems

1. Sep 30, 2008

a18c18

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

The speed of the head of a redheaded woodpecker reaches 3.7 m/s before impact with the tree. The mass of the head is 0.060 kg and the average force on the head during impact is 6.1 N.

(a) Find the acceleration of the head (assuming constant acceleration).
km/s2

(b) Find the depth of penetration into the tree.
cm

(c) Find the time it takes for the head to come to a stop.

2. Relevant equations

F=ma
F=mv^2/2d

3. The attempt at a solution

For a I tried to used the formula F=ma and divided the force by the mass to find the acceleration or 6.1/0.060 but this did not work.

I got the answer for b by using F=mv^2/2d and got 6.7cm which was correct.

In order to get the answer for c I think I need the answer for a.

2. Sep 30, 2008

Lazyshot

Are you sure for answer a that you are using the right units for the answer. km/s^2?

3. Sep 30, 2008

a18c18

I tried 6.1/.060 and got 101.6667 m/s^2. Then I divided 101.6667 by 1000 since the answer needs to be km/s^2 and got .1016 but neither of these answer were correct.

4. Sep 30, 2008

LowlyPion

Is that really km/s2?

Also the acceleration is really deceleration to zero. Perhaps you have the wrong sign?

Part b) is found by the V2/(2*a) = x relationship.

Part c) is found from x = 1/2 a*t2

5. Sep 30, 2008

a18c18

Okay I switched the signs for the acceleration and -.1016 was correct so thank you very much LowlyPion! but I still could not get c to work. Is x in c supposed to be the answer from b? I tried 6.7=.5*-.1016t^2 but got the wrong answer.

6. Sep 30, 2008

Lazyshot

Make sure your units match. You are putting km/s^2 in for a, but the putting distance x in cm. Either change a or x to match the other's units.

7. Sep 30, 2008

LowlyPion

Yes the answer from b is the distance. But whoa. wait. You have km/s2 acceleration and cm of distance. Please resolve that.

8. Sep 30, 2008

Lazyshot

Also the equation needs to have initial Velocity added to it:
X = V0*t + .5*a*t^2
Am I right?

9. Sep 30, 2008

LowlyPion

If it's uniform deceleration forget the initial velocity. Treat it as accelerating from rest. It's the same time over that distance.