Chapter 3.4, Problem 48E

To determine

To calculate:the time take by airplane to become airborne.

Answer to Problem 48E

Airplane will take $25\text{seconds}$ and travel distance of $694.44\text{meters}$ on runway to become airborne.

Explanation of Solution

Given information:

1. Distance traveled by air craft along a runway is defined by following function

  $D=\left(\frac{10}{9}\right)t^2$

Where $D$ is distance from starting point in meters and $t$ is time in seconds.

2. To become airborne speed of plane must reach to $200\text{km}/hr$

Formula used:

Speed of an object is first derivative of distance travelled with respect to time and is defined by following expression

  $v=\frac{d}{dt}D$

Calculation:

Functiongoverning distance travelled by airplane on runway is given by following expression.

  $D=\left(\frac{10}{9}\right)t^2$

Evaluate expression for speed by differentiation $D$ with respect to $t$ .

  $v=\frac{d}{dt}D$

  $=\frac{d}{dt}\left(\left(\frac{10}{9}\right)t^2\right)$

  $=\frac{10}{9}\frac{d}{dt}{t}^2$

  $=\frac{10}{9}\times 2t$

  $v=\frac{20t}{9}$

As distance covered is in meters and time is in seconds need to covert required speed $200\text{km}/hr$ in meter per second.

  $200\text{km}/hr=\frac{200000\text{meters}}{3600\text{seconds}}$

  $=\frac{500\text{meters}}{9\text{seconds}}$

  $=\frac{500}{9}\text{meters}/\sec$

Calculate time required by airplane to attain speed of $\frac{500}{9}\text{meters}/\sec$ .

  $v=\frac{20t}{9}$

  $\frac{500}{9}=\frac{20t}{9}$

  $t=\frac{500}{20}$

  $t=25\text{seconds}$

Calculate distance travelled by airplane in 25 seconds.

  $D=\left(\frac{10}{9}\right)t^2$

  $=\left(\frac{10}{9}\right)\times {25}^2$

  $=\frac{10\times 625}{9}$

  $=\frac{10\times 625}{9}$

  $=694.44\text{meters}$

Hence airplane will take $\underset{\_}{25\text{seconds}}$ and travel distance of $\underset{\_}{694.44\text{meters}}$ on runway to become airborne.