Question

Question: On average, an eye blink lasts about 100 ms. How far does a MiG - 25“Foxbat” fighter travel during a pilot’s blink if the plane’s average velocity is3400 km/hr?

Short answer

The distance traveled by the fighter plane in the given time is $94.44\text{m}$.

Step-by-step solution

Given data 

Average velocity,$v_{avg}=3400\text{km/h}$

Time interval, ${\text{∆t=100m/s=100×10}}^{-3}\text{s}$

Understanding the average velocity Average velocity is defined as the ratio of total displacement to the total time interval.  

The expression for the average velocity is given as follows:

$\left(v\right)=\frac{dis\tan ce\left(∆s\right)}{time\left(∆t\right)}$ $v_{avg}=\frac{∆x}{∆t}$ … (i)

Here, $∆x$ is the displacement and$∆t$ is the time interval

Determination of the distance traveled during the blink.

Convert the velocity from km/h to m/s as:

$$\begin{gathered} v=3400\frac{\text{km}}{\text{h}}\times \frac{1000\text{m}}{\text{km}}\times \frac{1\text{h}}{3600\text{s}} \\ =944.44\text{m/s} \end{gathered}$$

Using equation (i) the distance traveled can be calculated as follows:

$$\begin{gathered} ∆x=v\times ∆t \\ =944.44{\text{m/s×100×10}}^{-3}\text{s} \\ =94.44\text{m} \end{gathered}$$

Thus, the distance travelled by the fighter plane in the given time is $94.44\text{m}$.