Cracking the AP Physics B Exam (2012 Edition) by Princeton Review

By Princeton Review

This can be a mobi-pdf conversion, and as such there's loads of additional whitespace. i'd attempt to eliminate a few of it and reupload later. you may as well obtain the retail mobi version.

If you must comprehend it, it’s during this e-book! Cracking the AP Physics B examination, 2012 version, includes: 

   • A complete evaluation of vectors, fluid mechanics, optics, atomic and nuclear physics, and more
   • Step-by-step thoughts for cracking even the hardest problems
   • Detailed motives for the free-response component to the exam
   • Updated strategies that mirror the AP attempt scoring change
   • 2 full-length perform exams with special motives

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Extra resources for Cracking the AP Physics B Exam (2012 Edition)

Sample text

The other Big Five equations can be derived from these two definitions and the equation = (v0 + v), using a bit of algebra. 10 An object with an initial velocity of 4 m/s moves along a straight axis under constant acceleration. Three seconds later, its velocity is 14 m/s. How far did it travel during this time? Solution. We’re given v0, ∆t, and v, and we’re asked for x. So a is missing; it isn’t given and it isn’t asked for, and we use Big Five #1: Note: This last equation could also have been written as That is, it’s okay to leave off the units in the middle of the calculation as long as you remember to include them in your final answer.

It says that, at time t = 0, the object’s velocity was v = 0. Over the first two seconds, its velocity increased steadily to 10 m/s. At time t = 2 s, the velocity then began to decrease (eventually becoming v = 0, at time t = 3 s). 5 s. 5 s on, the velocity remained a steady −5 m/s. What can we ask about this motion? First, the fact that the velocity changed from t = 0 to t = 2 s tells us that the object accelerated. The acceleration during this tme was Note, however, that the ratio that defines the acceleration, ∆v/∆t, also defines the slope of the v vs.

For example, what if you wish to find the instantaneous velocity at 10 seconds? The velocity from 9−10 seconds is close to the velocity at 10 seconds, but it is still a bit too slow. The velocity from 10−11 seconds is also close to the velocity at 10 seconds, but it is still a bit too fast. You can find the middle ground between these two ideas, or the slope of the line that connects the point before and the point after 10 seconds. This is very close to the instantaneous velocity. A true tangent line touches the curve at only one point, but this line is close enough for our purposes.

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