# Solution for quadratic formula in physics problems [duplicate]

I’ve solved a physics problem about acceleration with the quadratic formula and I don’t understand the solutions.
There is two vehicles A and B. A is before B of 50 meters. The velocity of A is 20 $$ms^{-1}\hat{i}$$ and B is 30 $$ms^{-1}\hat{i}$$.Then A start an acceleration of 2 $$ms^{-2}$$. How long do A overtake B.
I’ve collapsed two equation such as:
$$\frac{1}{2}a_{A}t^2+v_{0A}t = x_{0B} + v_{B}$$
Then i have à second degree equation to find the time, and i solve it with the quadratic formula:
$$x_{1,2} = \frac{-b \pm \sqrt{b^2- 4ac}}{2a}$$
But then i have 2 solutions with one negative ex: 6 and -2. The positive solution is right but the negative can’t work because it lend to a negative time. Why there is a « false » solution ?

• The right side of your equation can't be correct as you are adding a distance and a velocity. The velocity term should be multiplied by t. When I solve this problem (assuming x=0 as initial position of vehicle A and t=0 at that time) I get 13.7 sec and -3.7 sec as solutions to the quadratic formula. Using the positive time, each vehicle is at a distance x~461 m at that time. Sep 20, 2020 at 15:12

The solution with negative $$t$$ has a physical meaning. It means that as well as the time in the future when the vehicles meet, there is also another time in the past (assuming all conditions stay the same) when the vehicles met. In other words, if we run time backwards starting from $$t=0$$ then the vehicles will meet at some negative time.

Suppose a ball is dropped from rest at a height of $$20$$ metres. When will it hit the ground (taking $$g$$ as $$10$$ m/s/s) ? We know that $$s = \frac 1 2 at^2 = 5t^2$$ so we can see that $$s=20$$ when $$t=2$$, and the downwards velocity of the ball at $$t=2$$ is $$v=20$$ m/s. But there is another solution: $$s=20$$ when $$t=-2$$. This means that if the ball were thrown up from the ground $$2$$ seconds ago with an upwards velocity of $$20$$ m/s then it would be at rest at a height of $$20$$ metres at $$t=0$$.

In these types of problems, we are usually only interested in what happens in the future - but the laws of classical physics are deterministic and extend backwards in time as well as forwards, so we can use the same equations and initial conditions to find what happened in the past as well.

Basically because the quadratic formula doesn't know the concept of time, we do and when we employ the formula to get solution it does its job and we get two solutions to the equation. Because we have defined time such that it can't be negative, we ignore the negative result. Say hypothetically time could be imagined to be negative, and try to retrace the paths of A and B you will find that after -2 seconds A will overtake B. This statement sounds absurd, I know and so negative solutions are ignored.

Surely what you have done is correct and when it comes to the elaboration of the steps that you take, solution to quadratic equation always yields 2 solutions (not necessarily different) either real or complex and in your physical system to determine which of the result is correct you need to check the physical boundaries.In your case you want to find time spend from an initial condition and for it you start time from 0s means you are excluding what happened before so you omit the negative result and if you begin with different initial conditions then you also need to consider them for the boundary of your solution. Also, considering the physical reality of your question negative result also means that at -2s cars were together so if you start the time counts from -2 then initially they were together.

As Butane mentioned, the negative solution has a physical meaning. Notice that car B always had a speed of 30, and A is accelerating but started with a smaller speed, and only reaches 20 at t=0. So the two cars would have meet in the past, assuming their accelerations have not changed.

• But OP stipulates that A starts accelerating at t=0. Yes, if it were always accelerating, then the negative solution would be meaningful to the problem. Sep 20, 2020 at 15:30

You are defining the acceleration of A to start at t=0. You then end up with a quadratic equation involving the acceleration of A. It therefore would be incorrect to use the negative t solution as before t=0 there was no acceleration so the equation couldn't apply to t<0.

Using the solution in my comment above, using the t=-3.7 sec solution we get x~-60 m for each vehicle. So mathematically speaking, t=-3.7 is a solution to the quadratic equation but I can't think of any physical meaning to that solution because A was not accelerating during t<0 so, again, the equation can't apply for those time periods.

Edit: see @Wolphram jonny's answer below.