# How to derive the equation with velocity and acceleration?

1. Feb 19, 2016

### Eclair_de_XII

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

2. Relevant equations
x - x0 = v0t + ½at2

3. 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

2. Feb 19, 2016

### haruspex

Compare this with s=s0+v0t+(1/2)at2. What have you forgotten?

3. Feb 19, 2016

### Eclair_de_XII

Let's see...

f(x) = s
x = t
A = a
B = v0
C = s0

so (assuming s0 = 0):

f(x) = -1.25x2 + 12x

Just to confirm...

28 = -1.25(4)2 + 12(4)
28 = 48 - 20
28 = 28

(b)
-1.25x2 + 12x = 2x + 20
-1.25x2 + 10x - 20 = 0
x2 - 8x + 16 = 0
x = ±4

So we can confirm that it's only once that car B matches car A at this acceleration.

(d)
At an arbitrarily higher acceleration...

-1x2 + 12x = 2x + 20
-1x2 + 10x - 20 = 0
x2 - 10x + 20 = 0

x = 5±√(5)

There are two points at which car B matches car A.

(e)
At an arbitrarily lower acceleration...

-1.26x2 + 12x = 2x + 20
-1.26x2 + 10x - 20 = 0
x2 - 7.397x + 15.873 = 0

The solution is imaginary, so at any lower acceleration, the cars do not cross paths.

4. Feb 19, 2016

### haruspex

That all looks right, except that I'm not sure you are entitled to choose specific higher and lower accelerations. The way I read the question, you should show that any higher (less negative) acceleration they meet twice, etc.

5. Feb 20, 2016

### Eclair_de_XII

Okay, I'm just going to do this very abstract-like, and you can tell me if I'm wrong or not.

Let a = arbitrary constant > 0

$(d)$
$(-1.25 + a)x^2+10x-20=0$
$x=\frac{-10±\sqrt{100+80(-1.25+a)}}{2(-1.25+a)}$
$x=\frac{-10±\sqrt{80a}}{2(-1.25+a)}$
$x=\frac{-10±4\sqrt{5a}}{2(-1.25+a)}$
$x=\frac{-5±2\sqrt{5a}}{(-1.25+a)}$

So that produces two distinct instances where cars A and B run into each other. And now for (e), which is basically the same thing, except you're subtracting a.

Let a = arbitrary constant > 0

$(e)$
$(-1.25 - a)x^2+10x-20=0$
$x=\frac{-10±\sqrt{100+80(-1.25-a)}}{2(-1.25-a)}$
$x=\frac{-10±\sqrt{-80a}}{2(-1.25+a)}$
$x=\frac{-5±2\sqrt{-5a}}{(-1.25+a)}$

Therefore, since there are no solutions to the system of equations for cars A and B, with an arbitrarily lower acceleration, they do not coincide at any point.

6. Feb 20, 2016

### Eclair_de_XII

Oh, damn, I just realized; I would have to replace (-1.25±a) with [(-2.5±a)/(2)] or (-1.25±a/2).

Well, whatever. I'm too lazy to fix it, and since (a/2) wouldn't affect the signs under the radicals, I guess I'm okay.