quadratic formula help for a physics problem

[gogogadget]

I had a question pertaining to physics but I am not sure I understand the steps my problem took it in setting the problem up as a quadratic equaton to solve it. I'll show the steps up to what I understand and how they solved it but they didn't show the specific algebraic steps to how they set up the quadratic equation itself. Many thanks for all who read this.

Problem:

A car traveling at a constant speed of 45.0 m/s (meters per second) passes a trooper on a motorcycle hidden behind a billboard. One second after the speeding car passes the billboard, the trooper sets out from the billboard to catch the car, accelerating at a constant rate of 3.00 m/s². How long does it take the trooper to overtake the car?

So for my equations I have:

v_xcar=45.0 m/s

t_car=1 second

a_xtrooper=3.00 m/s²

Because the car is traveling at a constant velocity, the equation for the constant velocity would be applied to find its final position at 1 second:

x=postion, t=time, v=velocity, a=acceleration

x_car=x_i + v_xcar

Because the car is moving at a speed of 45.0 m/s, it can be presumed that x_icar (initial position) is 45 m when it reaches the sign if t=0 upon passing the sign. Therefore, x_{car=45 m.}


In looking at the trooper, the trooper starts from rest at t_trooper=0 s. It accelerates at a_xtrooper=3.00 m/s². As such, it's position in moving at a constant acceleration can be calculated as shown below:

x_f=x + v_xi + 1/2a_xt² = (0) (0) = 1/2a_xt²

Therefore: x_trooper=x_car

1/2a_xt² = x_i + v_xcart

1/2a_xt^{2 -} v_xcart - x = 0

^{From here is where I get confused as the next step they show as:}



t = v_xcar +/-

v²_xcar + 2ax(x)/a_x



So I know that they factor out t and get everything else over to one side. I know they take the square root of t to get it on the other side. I don't understand though why there are two "ax's" though. There's the one that is multiplied by 2 and x and then the other that is divided. If someone could show me the step by step process in getting the square root equation for t set up, I'd really appreciate it as my book did not do it and I'm confused. Thank you again!

 

[MarkFL]

They are simply using the quadratic formula, where \(\displaystyle a=\dfrac{1}{2}a_x,\,b=-v_{x_{\text{car}}},\,c=x_{i_{\text{car}}}\).

 

[HallsofIvy]

Alternatively, complete the square:
\(\displaystyle (1/2)a_xt^2- v_{xcar}t- x= 0\) can be written as \(\displaystyle (1/2)a_xt^2- v_{xcar}t= x\). Then, dividing through by \(\displaystyle (1/2)a_x\), \(\displaystyle t^2- (2v_{xcar}/a_x)t= 2x/v_{xcar}\).


Now I presume you know that \(\displaystyle (t- a)^2= t^2- 2at+ a^2\). Comparing that to the left side of the previous equation, we can see that \(\displaystyle 2a= 2v_{xcar}/a_x\) so \(\displaystyle a = v_{xcar}/a_x\) and \(\displaystyle a^2= (v_{xcar}/a_x)^2\) that is, if we add \(\displaystyle (v_{xcar}/a_x)^2\) to both sides we have \(\displaystyle t^2- (2v_{xcar}/a_x)t+ (v_{xcar}/a_x)^2= (t- v_{xcar}/a_x)^2= x+ (v_{xcar}/a_x)^2\(\displaystyle .

Taking the square root of both sides, \(\displaystyle t- v_{xcar}/a_x= \pm\sqrt{x+ (v_{xcar}/a_x)^2}\) and so \(\displaystyle t= v_{xcar}/a_x\pm\sqrt{x+(v_{xcar}/a_x)^2}\).\)\)

 

[JeffM]

So I know that they factor out t and get everything else over to one side. I know they take the square root of t to get it on the other side. I don't understand though why there are two "ax's" though. There's the one that is multiplied by 2 and x and then the other that is divided. If someone could show me the step by step process in getting the square root equation for t set up, I'd really appreciate it as my book did not do it and I'm confused. Thank you again!

No wonder you are confused. I think that is a very confusing way to attack the problem. Let's start with more enlightening definitions:

\(\displaystyle x_{final} = distance\ from\ sign\ when\ chase\ ends;\)

\(\displaystyle x_{initial\ car} = distance\ of\ car\ from\ sign\ when\ chase\ starts;\ and\)

\(\displaystyle x_{initial\ trooper} = distance\ of\ trooper\ from\ sign\ when\ chase\ starts.\)

With those definitions it is clear why:

\(\displaystyle Equation\ \alpha:\ x_{initial\ car} = 45.\) Clear defintions make thinking easy.

\(\displaystyle Equation\ \beta:\ x_{initial\ trooper} = 0.\)

\(\displaystyle Equation\ \gamma:\ x_{final} = x_{initial\ car} + 45t + \frac{1}{2}(0)t^2 = 45 + 45t + 0 = 45 + 45t.\) Simplified using equation alpha.
\(\displaystyle Equation\ delta:\ x_{final} = x_{initial\ trooper} + (0)t + \frac{1}{2}(3)t^2 = 0 + 0 + \dfrac{3t^2}{2}= \dfrac{3t^2}{2}.\) Simplified by equation beta.

Nothing is factored out, especially not t. After simplifying equations gamma and delta by using equations alpha and beta, they are EQUATED to get:

\(\displaystyle Equation\ \epsilon:\ 45 + 45t = \dfrac{3t^2}{2} \implies 90 + 90t = 3t^2 \implies 3t^2 - 90t - 90 = 0.\)

As has been explained before, this can be solved using the quadratic formula:

\(\displaystyle 3t^2 - 90t - 90 = 0 \implies t^2 - 30t - 30 = 0 \implies t = \dfrac{-(-30) \pm \sqrt{(-30)^2 - (4)(1)(-30}}{2 * 1} = \dfrac{30 \pm \sqrt{900 + 120}}{2} =\)

\(\displaystyle \dfrac{30 \pm \sqrt{1020}}{2} = 15 \pm \sqrt{\dfrac{1020}{2^2}} = 15 \pm \sqrt{255}.\)

Very straight forward. Here is what the book did.

It got: \(\displaystyle \left(\frac{1}{2} * a_{trooper} * t^2\right) - \left(v_{car} * t\right) - x_{initial\ car} = 0.\)

If we now substitute numbers in for the variables, that translates into: \(\displaystyle \frac{1}{2}(3t^2) - 45t - 45 = 0.\)

If, to get rid of the 1/2, you then multiply both sides of that equation by 2, you get: \(\displaystyle 3t^2 - 90t - 90 = 0.\)

But that is exactly what I got. In other words, they are not replacing variables with numeric values until the very end, but the logic is exactly the same as mine. Then, still keeping variables and without getting rid of the fraction, they apply the quadratic formula:

\(\displaystyle t = \dfrac{- (- v_{car}) \pm \sqrt{(- v_{car})^2 - 4\left(\frac{1}{2} * a_{trooper}\right)(x_{initial\ car})}}{2 * \left(\frac{1}{2} * a_{trooper}\right)} \implies\)

\(\displaystyle t = \dfrac{v_{car} \pm \sqrt{(v_{car})^2 + (2 * a_{trooper} * x_{initial\ car})}}{a_{trooper}}.\)

Again substituting numeric values, the book gets:

\(\displaystyle t = \dfrac{45 \pm \sqrt{45^2 + (2 * 3 * 45)}}{3} = \dfrac{45 \pm \sqrt{2025 + 270}}{3} = \dfrac{45 \pm \sqrt{2295}}{3} = 15 \pm \sqrt{\dfrac{2295}{3^2}} = 15 \pm \sqrt{255}.\)

Same answer as mine. I do not like the book's notation, nor do I like its general presentation, but there is nothing wrong with its logic.

Does this help?