Chapter 1.5, Problem 36E

To determine

To explain:

The time by each flight

Answer to Problem 36E

The time by first flight and second flight is $2.5hrs$ and $2.3hrs$ respectively.

Explanation of Solution

Given information:

  $\begin{array}{l}{v}_1=460mph\\ v_2=500mph\\ t=4.8hrs\end{array}$

Formula used:

The average velocity of flight is given by the following expression:

  $v=\frac{d}{t}$

Calculation:

Consider,

Time taken first flight: $t_1$

Time taken by second flight: $t_2$

Now, total time taken for flying is $t=4.8\sec$ :

  $\begin{array}{l}t=4.8\sec \\ t_1+t_2=4.8\end{array}$

By using the formula, get the value of the distance covered by the flight:

  $\begin{array}{l}{t}_1+t_2=4.8\\ \frac{d}{v_1}+\frac{d}{v_2}=4.8\end{array}$

Simplify:

  $\frac{d{v}_2+d{v}_1}{v_1{v}_2}=4.8$

Multiply by $v_1{v}_2$ :

  $\begin{array}{l}\frac{d{v}_2+d{v}_1}{v_1{v}_2}\times v_1{v}_2=4.8\times v_1{v}_2\\ d\left(v_2+v_1\right)=4.8\times v_1{v}_2\end{array}$

Divide by $\left(v_1+v_2\right)$ :

  $\begin{array}{l}d\frac{\left(v_2+v_1\right)}{\left(v_2+v_1\right)}=4.8\times \frac{v_1{v}_2}{\left(v_2+v_1\right)}\\ d=4.8\times \frac{v_1{v}_2}{\left(v_2+v_1\right)}\end{array}$

Now, the time taken by first flight is calculated as:

  $\begin{array}{l}{t}_1=\frac{d}{v_1}\\ t_1=\frac{4.8\times \frac{v_1{v}_2}{\left(v_2+v_1\right)}}{v_1}\\ t_1=4.8\times \frac{v_2}{\left(v_2+v_1\right)}\\ t_1=4.8\times \frac{500}{\left(500+460\right)}\\ t_1=2.5hrs\end{array}$

Now, the time taken by first flight is calculated as:

  $\begin{array}{l}{t}_1+t_2=4.8\\ 2.5+t_2=4.8\\ t_2=2.3hrs\end{array}$