## Problem Statement

**An engineer is designing the runway for an airport. ****Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s ^{2}. **

**The takeoff speed for this plane will be 65 m/s.**

**Assuming this minimum acceleration, what is the minimum allowed length for the runway?**

## Solution

**U = 0 V = 65 m/sa = 3 m/s**

^{2}

**To find out the distance, we will use the following equation:S = Ut + ½ a t ^{2}as U = 0, the equation becomes S = ½ a t^{2} = ½ . 3. t^{2} ……(1)**

**We will use the following eqn to find out the time required to attain this velocity. V = u + atAs U = 0, the eqn becomes V = att = V/a = 65/3 sec……(2)**

**Putting the value of time t in equation 1 we get S = ½ . 3. (65/3) ^{2}= 704 mSo the minimum allowed length of the airport is 704 m**