MHB How Can I Correctly Derive the Quadratic Equation from a Train Speed Problem?

[Simonio]

I'm having difficulty deriving a quadratic from the info in this question:

A train usually covers a journey of 240 km at a steady speed of \(v \text{ kmh}^{-1}\). One day, due to adverse weather conditions, it reduces its speed by 40 \text{kmh}^{-1}\) and the journey takes an hour longer.

Derive the equation \(v^2 - 40v - 9600 =0\), and solve it to find the value of \(v\).

All I could come up with was: \(\frac{240}{v^2} = \frac{240}{v^2 - 40} -1\)

But i don't think this is right! Any help to get me on the right track appreciated! Thanks.

 

[MarkFL]

Well, let's begin with the kinematic relationship between distance $d$, speed $v$, and time $t$, when speed is constant:

$$d=vt$$

Let's let the distance be measured in km, speed in kph and time in hrs.

We know in both cases that $d=240$. Under usual conditions, we are given:

$$240=vt$$

And when the weather was bad, we are given:

$$240=(v-40)(t+1)$$

Now, we want to get rid of $t$ so that we have an equation in $v$ only. Solving the first equation for $t$, we obtain:

$$t=\frac{240}{v}$$

Now, substitute for $t$ into the second equation:

$$240=(v-40)\left(\frac{240}{v}+1\right)$$

Can you continue?

 

[Simonio]

Well, let's begin with the kinematic relationship between distance $d$, speed $v$, and time $t$, when speed is constant:

$$d=vt$$

Let's let the distance be measured in km, speed in kph and time in hrs.

We know in both cases that $d=240$. Under usual conditions, we are given:

$$240=vt$$

And when the weather was bad, we are given:

$$240=(v-40)(t+1)$$

Now, we want to get rid of $t$ so that we have an equation in $v$ only. Solving the first equation for $t$, we obtain:

$$t=\frac{240}{v}$$

Now, substitute for $t$ into the second equation:

$$240=(v-40)\left(\frac{240}{v}+1\right)$$

Can you continue?

Do we have to multiply throughout by \(v\)? Would that look like this:

\(v(v-40)v\) \((\frac{240}{v}+1)\)

Hmm...think I've gone wrong somewhere...

 

[MarkFL]

Do we have to multiply throughout by \(v\)? Would that look like this:

\(v(v-40)v\) \((\frac{240}{v}+1)\)

Hmm...think I've gone wrong somewhere...

Yes, multiplying through by $v$ is a good next step, and you would have:

$$240v=(v-40)(240+v)$$

Do you see what I did there?

 

[Simonio]

Yes, multiplying through by $v$ is a good next step, and you would have:

$$240v=(v-40)(240+v)$$

Do you see what I did there?

Thanks. Could you show the working..so I know how you got there? Thanks.

 

[Deveno]

We start with:

$240 = (v - 40)\left(\dfrac{240}{v} + 1\right)$

If we multiply one side by $v$, we must do the same to the other side.

The left-hand side is easy: we wind up with $240v$.

The right-hand side is a bit more involved:

$(v - 40)\left(\dfrac{240}{v} + 1\right)v$

$ = (v - 40)\left(\dfrac{240}{v}\cdot v + v\right)$ (using the DISTRIBUTIVE rule)

$ = (v - 40)(240 + v)$ ("cancelling the $v$'s")

$ = v(240 + v) - 40(240 + v)$ (distributive rule again, expanding out this time)

$ = 240 v + v^2 - 9600 - 40v$ ( another expansion using the distributive rule)

$ = v^2 + 200v - 9600$ (collecting "like terms", and ordering by the power of $v$).

So that's what we wind up with the right-hand side, and now we equate the two sides:

$240v = v^2 + 200v - 9600$

Do you see what to do next?

 

[Simonio]

So:

\(v^2 + 200v - 9600 =240v\)

Then: \(v^2 - 40v -9600 = 0\)

next, find factors of \(ac\) whose sum is \(c\) = 120, -80.

\(v2 + 120v -80v - 9600\)

Then: \(v(v-80) + 120(v-80)\)

So either \(v+120 = 0\) or \(v-80 = 0\)

Then \(v = -120\) or \(v = 80\)

hmm... the answer in my book gives v as +120 only-so I've gone wrong somewhere(Doh)

 

[anemone]

Since we have $v^2-40v-9600=0$:

\(v2 + 120v -80v - 9600\)

The above should be $v^2-120v+80v-9600=0$...:) and try it again, Simonio, as I'm fairly certain you will end up getting $v=120$ as the positive solution to the problem.

 

[Deveno]

So:

\(v^2 + 200v - 9600 =240v\)

Then: \(v^2 - 40v -9600 = 0\)

next, find factors of \(ac\) whose sum is \(c\) = 120, -80.

\(v2 + 120v -80v - 9600\)

Then: \(v(v-80) + 120(v-80)\)

So either \(v+120 = 0\) or \(v-80 = 0\)

Then \(v = -120\) or \(v = 80\)

hmm... the answer in my book gives v as +120 only-so I've gone wrong somewhere(Doh)

What you have written is tantamount to:

$v^2 - 40v - 9600 = (v - 80)(v + 120)$.

However, in truth:

$(v - 80)(v + 120) = v^2 + 40v - 9600$

as -80 and 120 sum to +40, not -40, as required.

In general, if $ab < 0$ (and $a,b$ are integers), then exactly ONE of $a$ or $b$ must be negative. If $a + b < 0$ as well, the negative one has to be the one with the largest absolute value (or magnitude).

If we write:

$v^2 - 40v - 9600 = (v + a)(v + b) = v^2 + (a+b)v + ab$

we have $ab = -9600 < 0$ and $a+b = -40 < 0$.