- #1
Eclair_de_XII
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Homework Statement
"Cars A and B move in the same direction in adjacent lanes. The position x of car A is given in Fig. 2-27, from time t = 0 to t = 7 s. The figure's vertical scaling is set by xs = 32 m. At t = 0, car B is at x = 0, with a velocity of 12 m/s and a negative constant acceleration aB. (a) What must aB be such that the cars are (momentarily) side by side (momentarily at the same value of x) at t = 4 s? (b) For that value of aB, how many times are the cars side by side? How many times will the cars be side by side if the magnitude of acceleration aB is (d) more than and (e) less than the answer to part (a)?"
t = 4 s
t(4) = 28 m
vB(0) = 12 m/s
Homework Equations
x - x0 = v0t + ½at2
The Attempt at a Solution
28 m - 0 m = (12 m/s)(4 s) + ½a(4 s)2
-20 m = ½a(16 s2)
-40 m = a(16 s2)
a = -2.5 m/s2
Right now, I'm working out part (d) and (e), assuming that for a = -2.5 m/s2, car B crosses car A's path only once. I'm trying to do this by way of deriving equations. For car A, I've derived an equation that describes its path: f(x) = 2x + 20. For car B, I'm trying to figure out how to express its motion (given by v0 = 12 m/s and a = -2.5 m/s2). This way, I can figure out how many times car B crosses car A, just by graphing the function for car B, and seeing how many intersections there are. I'm thinking the first coefficient of this equation would be f(x) = -2.5x2 + 12x+ C... but I wouldn't be sure how to derive the rest of the equation. Here's a picture of the graph from which I derived my first function:
http://i.imgur.com/O80znQM.jpg