Q110P - Fundamentals Of Physics

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 is  3400 km/hr?

Step-by-Step Solution

Answer

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

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 $94.44\text{ m}$ .