How Do You Calculate Time and Acceleration in Physics Problems?

[Tentothe]

I'm having some trouble with a couple more problems. Any help is appreciated.

Homework Statement

(a) An indestructible bullet 1.90 cm long is fired straight through a board that is 10.3 cm thick. The bullet strikes the board with a speed of 426 m/s and emerges with a speed of 278 m/s. What is the total time that the bullet is in contact with the board?

(b) A woman is reported to have fallen 34 ft from the top of a building, landing on a metal ventilator box, which she crushed to a depth of 25 in. Calculate her average acceleration while in contact with the box.

Homework Equations

(i) $$\Delta x = v_{o}t + \frac{1}{2}at^{2}$$

(ii) $$a = \frac{v^{2} - v^{2}_{o}}{2\Delta y}$$

The Attempt at a Solution

(a) I actually stumbled upon the answer for this one, but I ended up with two different values for $$t$$ and I'm wondering what the incorrect one means exactly.

For this problem, I first noted that the bullet is in contact with the board for both the thickness of the board (10.3 cm) as well as for the length of the bullet at both ends of the board as it exits and enters (2 * 1.90 cm). I had previously calculated the average acceleration of the bullet as it passed through the board and used that value along with this displacement and the given velocity and plugged them into equation (i) to find the time taken to pass through the board. Thus,

$$v_{o}$$ = 426 m/s, $$\Delta x_{}$$ = [10.3 + (1.90)(2)]/100 = 0.122 m, $$a$$ = -505786.4 m/s$$^{2}$$ (calculated and verified previously)

Plugging these into equation (i) and solving for $$t$$ using the quadratic formula gave me $$t$$ = 0.000366 s, 0.00132 s. The first turned out to be the correct answer, but now I'm confused as to what the second represents. Is there an easier way for calculating this time that I'm not seeing?

(b) I had previously calculated the woman's velocity just before hitting the box and used that with the displacement of 25 in. to calculate the average acceleration using equation (ii) as such:

$$v_{o}$$ = 46.76 ft./s (calculated and verified previously), $$\Delta y$$ = 25 in. * (1 ft./12 in.) = 2.083 ft., $$v_{f}$$ = 0 ft./s

$$a$$ = [(0)$$^{2}$$ - (46.76)$$^{2}$$]/(2 * 2.083) = -524.8 ft./s$$^{2}$$

Where have I gone wrong? Thanks.

 

[Kurdt]

For (a) I would have used the formula:

$$s = \left(\frac{u+v}{2}\right) t$$

For (b) I can't really see anything wrong.

 

[Moetasim]

For problem (a), the second value for t that you calculated is likely an extraneous solution. When using the quadratic formula, it is possible to get two values for t, but only one of them will be the correct answer. In this case, the first value of t that you calculated is the correct one, as it makes sense in the context of the problem.

For problem (b), you have mixed up the values for v and v_o in equation (ii). The initial velocity, v_o, should be 0 ft./s, as the woman is initially at rest before falling. The final velocity, v, should be 46.76 ft./s, as that is the velocity she reaches just before hitting the box. The correct calculation should be:

a = [(46.76)^2 - (0)^2]/(2 * 2.083) = 524.8 ft./s^2

Remember, in this equation, the initial velocity is squared and the final velocity is not. This is because the woman starts at rest and accelerates to a final velocity of 46.76 ft./s.