What is total distance travelled by plane on two different speeds.

Question

I was trying to solve a question today.

An aircraft travels at a speed of 180mph during the first half and at a speed of 460mph during the second half of its route. Total flight time is 04:15 hours. What is the distance between airport of departure and arrival.

Now the answer to this is 1100 miles.Would anyone mind to help me understand how we can get to this answer? The final formula would be: x = v1*t - (v1 - v2) * (v1t/(v1+v2)) The solution involved 42 lines of rewrites of formulas, but I am trying to understand it conceptually without having to rewrite it all.

Answer 1

$t_1 = \frac{x}{2v_1}$

$t_2 = \frac{x}{2v_2}$

$t_1 + t_2 = \frac{17}{4}$

You know $v_1$ and $v_2$ Find x comes to $1099.6875 \approx 1100$ miles

$\frac{x}{2v_1}+\frac{x}{2v_2} = \frac{17}{4}$

$ x\left(\frac{v_1+v_2}{2v_1v_2}\right) =\frac{17}{4}$

$x = \frac{180\times460\times17\times2}{640\times4} = 1100$

Answer 2

We have: $$ \frac d2\left(\frac 1 {v_1}+ \frac 1 {v_2}\right)=t\implies d=\frac {2v_1v_2}{v_1+v_2}t.$$

Answer 3

You have the speeds $v_1$ and $v_2$ on each of the two parts of the route, you have the total flight time $t$ for both parts of the route, you have that the distance of each part of the route was half the total distance, and you want to know the total distance $x.$

So if the times on the first and second parts were $t_1$ and $t_2$ and the distances were $x_1$ and $x_2$ respectively, then \begin{align} x_1 &= x_2 = \frac x2, \tag1\\ x_1 &= v_1 t_1,\tag2\\ x_2 &= v_2 t_2,\tag3\\ t &= t_1 + t_2.\tag4 \end{align}

We have $\frac x2$ in Equation $(1)$ because we were told we had a first half and a second half of the route.

Now from Equations $(2)$ and $(3)$ we have \begin{align} t_1 &= \frac{x_1}{v_1},\tag5\\ t_2 &= \frac{x_2}{v_2}.\tag6 \end{align}

Use Equation $(1)$ to substitute for $x_1$ and $x_2$: \begin{align} t_1 &= \frac{x/2}{v_1},\tag7\\ t_2 &= \frac{x/2}{v_2}.\tag8 \end{align}

Add Equations $(7)$ and $(8)$ to get $$ t_1 + t_2 = \frac{x/2}{v_1} + \frac{x/2}{v_2} = \frac x2\left( \frac{1}{v_1} + \frac{1}{v_2} \right). \tag9$$

Recalling that $t_1 + t_2 = t,$ $$ t = \frac x2\left( \frac{1}{v_1} + \frac{1}{v_2} \right). \tag{10}$$

Solve for $x$: $$ x = \frac{2t}{\frac{1}{v_1} + \frac{1}{v_2}}. \tag{11} $$

Technically, that's a completely solution. Just plug in the known data on the right hand side of Equation $(11)$ and compute. But we can clear the fractions in the denominator on the right-hand side of $(11)$ by multiplying by $v_1$ and multiplying by $v_2$: $$ x = \frac{2v_1v_2t}{v_2 + v_1}. \tag{12} $$

That's the same result as both the other answers, just derived in more detail; a lot more detail that most people would give. For example, you could start with Equations $(4),$ $(7),$ and $(8)$ and work from there.

Now let's take a look at the right-hand side of the solution you were given:

$$v_1 t - (v_1 - v_2) \left(\frac{v_1 t}{v_1 + v_2}\right). $$

Multiply $v_1 - v_2$ into the numerator of the fraction on the right,: $$v_1 t - (v_1 - v_2) \left(\frac{v_1 t}{v_1 + v_2}\right) = v_1 t - \frac{(v_1 - v_2) v_1t}{v_1 + v_2}. $$

Put everything over a common denominator: $$v_1 t - \frac{(v_1 - v_2) v_1t}{v_1 + v_2} = \frac{(v_1 + v_2)v_1 t - (v_1 - v_2) v_1t}{v_1 + v_2} .$$

Simplify: \begin{align} \frac{(v_1 + v_2)v_1 t - (v_1 - v_2) v_1t}{v_1 + v_2} &= \frac{(v_1^2 t + v_2v_1 t) - (v_1^2 t - v_2v_1t)}{v_1 + v_2}\\ &= \frac{v_1^2 t + v_2v_1 t - v_1^2 t + v_2v_1t}{v_1 + v_2}\\ &= \frac{2v_1v_2 t}{v_1 + v_2}. \end{align}

Notice that this is the same as the right-hand side of Equation $(12).$ So the solutions agree although they use different sequences of operations in their derivations.

The actual sequence of operations on the solution sheet looks like someone almost randomly trying to combine the equations in many different ways until something clicks. This is a legitimate thing to try if you don't have any better idea how to solve a problem, but it's a bit strange to see this from someone who (one would think) is supposed to be teaching the material.

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Here's an approach that I find a bit more intuitive. Notice that the speed on the second half of the route is $\frac{460}{180} = \frac{23}{9}$ times as fast as the speed on the first half. That means the second half of the route takes $\frac{9}{23}$ times as long a time to fly as the first half, because the distances were equal and time is inversely proportional to speed over a given distance.

That also means the two parts of the route divide the total time period in the ratio $23 : 9$ (to make the second part $\frac{9}{23}$ as much as the first part). Imagine dividing the total time into $32$ equal parts; then you can use $23$ of them to fly the first half of the route and $9$ to fly the second half. So you spend $\frac{9}{32}$ of the total time flying the second half of the route. Now consider that the total flying time is $4.25$ hours, that means you spend $$ \frac{9}{32} \times 4.25 = 1.1953125 $$ hours on that half of the route. During that time you fly at $460$ miles per hour, so the distance covered is $$460 * 1.1953125 = 549.84375 $$ miles. Multiply by $2$ to get the length of the entire route and not just one half of it.