# I have set values for $x$,$x_o$, $v$, $v_o$; I can't seem to get the same “$a$” value (kinematics equation) [closed]

I hope I can get a hand here. This seems stupidly simple, but I'm getting different answers depending on the equation I use.

I've read a word problem with values: \begin{align} t& = 0.200 s & x &= 200 m & x_0& = 0.00 m\\ v& = 0.00 m & v_0& = 40.0 m \end{align}

I'm expected to solve for "$a$".

I'm using three of the four classic kinematics equations, (because the fourth does not involve "$a$"), and will list them here if necessary. I can also repeat the entire question.

My problem comes when solving for "$a$": I get $-200 m/s^2$ when solving using one equation, and $9600 m/s^2$ using another, and yet another yields $-4.00 m/s^2$.

It is driving me pretty crazy. Thanks for taking the time to read.

Edit: The details are: A car suddenly brakes in order to stop at a stop sign. To stop fully it takes $200 m$ of distance from the time the brakes are applied, as well as $0.200 s$.

Additionally, the car was going $40.0 m/s$ in the instant it began to stop. This information is the source of my above values.

The formulas used for solving to find "$a$" are: \begin{align} (1) v& = v_0 + at & (2) x &= x_0 + v_0t + \frac{1}{2}at^2\\ (3) v^2&= v_0^2+ 2a(x-x_0) & (4) x &= x_0+ \frac{1}{2}(v+v_0)t \end{align}

Thanks to DelCrosB for the request to clarify. Thanks once more to Unnikrishnan for the formatting.

## closed as off-topic by ACuriousMind♦, Jon Custer, John Rennie, user36790, GertSep 21 '16 at 2:16

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• it is probably better if you give us the full text of your problem and the equations that you tried to use to solve it. What you have written so far is quite cryptic... – DelCrosB Sep 19 '16 at 18:41
• Looks like the anszats $v(t)=v(0)+a\cdot t$ and $x(T)=x_0 + \int_{0}^{T} v(t)dt$ leads you to the right acceleration according to the answer below, thinking in terms of integrals instead of premade formulas might be good for the memory. – Emil Sep 19 '16 at 23:04
• Emil, it's a little late for me to try and understand this equation in the context of this question (I was never explicitly given v(t); should I integrate a(t)?) but I will attempt to grasp it. Unfortunately, I have been the victim of cookie-cutter formulas and dumbed down exercises. As much as I love calculus, I have used very little in my Uni Physics. Can you recommend a text book or source? And, further, could you recommend somewhere to use it? Odd question, but as an undergrad I wonder where I'd actually end up using such equations/knowledge, if not in class. – CppWiz Sep 20 '16 at 4:08
• I like Mathematical Physics by Sadri Hassani and Manifolds, Tensor Analysis, and Applications by Marsden, Jerrold E. and Ratiu, Tudor and Abraham, Ralph. I don't understand everything in them and I'm not sure they cover calculus but they have a lot of different things in them. – Emil Sep 20 '16 at 5:40
• Sure, you can use it to answer questions like this on this site. – Emil Sep 20 '16 at 5:41

## 3 Answers

I think there is either an error in the time of 0.200s given, or the meaning of that time is being misunderstood. I don't see how it can be meant as the time it takes to come to a stop from the instant the brakes engage. It's simply too short. It can't make sense if the stopping distance of 200m is correct.

I feel your acceleration value of -4 meters per second squared must be what was intended. If so, then the time to stop would be 10 seconds, not 0.2.

Is it possible the problem may have meant that 0.2s was some sort of "reaction time" between when the driver decided to stop and the actual pressing of the brake? Are you sure you gave a verbatim statement of the original problem above?

All of the equations are going to vary. Number 1 does not take distance into account, number 2 does not take final velocity into account, and the third does not take time into account.

For you to get this problem correct you need to use equation 1.

In any kinematics problem involving constant acceleration, there are five variables: initial velocity, final velocity, acceleration, time, and displacement.

Further, there are five equations that state relationships between these variables. These are (seen in https://en.wikipedia.org/wiki/Equations_of_motion)

Note that each equation:

1. uses four of the variables and ignores one;
2. Each of the five variables is ignored by one and only one of the equations.

So, for any particular problem, if you are given exactly three variables, you can find two equations that use these three variables; one can be solved for the fourth variable, the other can be solved for the fifth variable.

For example, if you are given $t, u, \text{ and } s$, you can use Equation [2] above to find $a$, or use Equation [3] to find $v$. Of course, some algebraic manipulation will usually be needed.

The difficulty with the OP's question is that there are already four variables given: in terms of the symbols used here, $t = 0.200$, $v=0$, $u=40$, $s=200$.

If you put these given values into Equation [3] above, the values do not satisfy the equation. The problem as stated is impossible. That's why the OP can't solve it...

• For a hand-waving objection, the car goes 200 m in 0.200 sec. That's averaging 1000 m/s. Yet it's supposedly slowing from 40 m/s to zero? – DJohnM Sep 20 '16 at 2:50