AP Physics

Test Preparation
1

<p>Hello everyone, I have just began my AP Phsyics course and I have some trouble with the following two problems and would appreciate it if someone would help:</p>

<li><p>The minimum distance to stop a car moving at 35 mi/hr is 40 ft. What is the minimum stopping distnace for the same car moving at 70 mi/hr, assuming the same rate of acceleration?</p></li>
<li><p>An indestructible bullet 2.00 cm long is fired straight through a board that is 10.0 cm thick. The bullet strikes the board with a speed of 420 m/s and emerges with a speed of 280 m/s. a. What is the average acceleration of the bullet through the board? b. What is the total time the bullet is in contact with the board? c. What thickness of the board would it take to stop the bullet (calculated to 0.1 cm)?</p></li>
</ol>

<p>Thanks for the help.</p>

2

<p>bump please</p>

3

<ol>
<li>40 ft* (mi/5280 ft) = 0.0076 mi
Vf^2 - Vo^2 = 2ax
0^2 - (35)^2 = 2a(0.0076)
80592 mi/h^2 = a</li>
</ol>

<p>x = (Vf^2 - Vo^2)/(2a)
x = (0^2 - (70)^2)/(2(80592))
x = 0.0304 mi or 160.5 ft</p>

<ol>
<li>a) Vf^2 - Vo^2 = 2ax
(280)^2 - (420)^2 = 2a(0.10)
-490,000 m/s^2 = a</li>
</ol>

<p>b) This is a little tricky to me given the length of the bullet and how to approach that aspect of the problem. Here's my best guess:</p>

<p>x = vt
0.12 = [(420 + 280)/2]t
3.429*10^-4 s = t</p>

<p>c) Assuming that the acceleration is the same,</p>

<p>x = (Vf^2 - Vo2)/(2a)
x = ((0)^2 - (420)^2)/(2(-490,000))
x = 0.18 m = 18.0 cm</p>

<p>That's my best stab at them.</p>

4

<p>come on please, I have to go to bed in 30 minutes</p>

5

<p>I take it that my answers are incorrect then?</p>

6

<p>"come on please, I have to go to bed in 30 minutes"
looks like you forgot to mention another problem, your attitude.</p>

<p>btw, you really shouldn't be using this site for hw help.</p>

7

<p>For 2b), I don't think that's how you should solve it. This is my solution for 2b).</p>

<p>subscript o indicates the value at the beginning</p>

<p>x = x<em>o + (v</em>o)t + (1/2)at^2
(0.120m) = 0 m + (420m/s)t + (1/2)(-490000m/s^2)t^2
t = (-420 +/- 58800)/(-490 000)
At this step, only (-420 - 58800)/(-490 000) is only valid since if you used the other one results in a negative value for time, which is not possible. Therefore, t = 0.121 s</p>

8

<p>hewashere- :D</p>

9

<p>Interesting. Shouldn't we be able to prove it that way, though?</p>

<p>If you use x = vt, and just pretend that v is constant and = 280 m/s during the .12m it has to travel, the SLOWEST time it can have is still 4.something x 10^-4 s.</p>

<p>Likewise, if you use x = vt again, 0.12 = v(0.121), you get an average velocity of 0.99 m/s, which doesn't even come close to the 280 to 420m/s range.</p>

<p>Just my $0.02.</p>

10

<p>Go to the Physics Forums</p>

11

<p>lol, first of all, i realized I did the calculation wrong the first time. I forgot to take the square root when I used to quadratic formula =P</p>

<p>Also, I realized something as I was reading your comment, setzwxman. Doesn't the acceleration only apply when the bullet is going through the board? meaning during the last 2 cm, there is no acceleration? So in that case, </p>

<p>(0.100m) = 0 m + (420m/s)t + (1/2)(-490000m/s^2)t^2
at this point, t = (- 420 +/- 280) / (-490000)</p>

<p>Now, since - 420 + 280 or -420-280 both leads to a negative answer, I used your logic to find out that -420+280 is much more plausible than -420+280. Therefore, t = (-420 + 280)/(-490000) = 2.86 x 10^-4</p>

<p>For the last 2.00cm,</p>

<p>x = vt
(0.02 m) = (280 m/s)t
t = 7.143 x 10^-5</p>

<p>Therefore, we add the two together, resulting in 3.57 x 10^-4</p>

12

<p>I am not sure, but even during the last 2cm, isn't the bullet still in contact w/ the board? If it is, I'd think that the acceleration still applies.</p>

13

<p>lol, then we shoulda used 12.0cm for part a also, shouldn't we?</p>

14

<p>I always thought when finding the acceleration or velocity (like in a lab situation or whatever), you are supposed to use only the FRONT portion of the object for time/length measurements. That's why I only used 0.10m for part a.</p>

<p>For part b, instead of sticking w/ the front, you have to add on another 0.02m to the length it has to pass w/ that acceleration value. I'm just speculating...I apologize if I am wrong.</p>