Question

An airplane starting from a stop accelerates down a runway at 3.20 m/s^2 for 32.8 seconds until it finally lifts off the ground. Determine the distance traveled before takeoff in m.

Answer 1

$1721.34\text{ meters }$

Explanation

to solve this we need to use the equation :

$\begin{gathered} x=v_0t+\frac{at^2}{2} \\ \text{where} \\ x\text{ is the traveled distance} \\ v_ois\text{ the initial sp}eed \\ a\text{ is the acceleration} \\ t\text{ is the time} \end{gathered}$

Step 1

Let

$\begin{gathered} v_0=0\text{ ( it starts from the stop)} \\ a=3.20\frac{m}{s^2} \\ t=32.8\text{ sec} \\ x=x \end{gathered}$

replace

$\begin{gathered} x=v_0t+\frac{at^2}{2} \\ x=(o)(32.8s)+\frac{(3.2\frac{m}{s^2})(32.8s)^2}{2} \\ x=0)+\frac{(3.2\frac{m}{s^2})(32.8s)^2}{2} \\ x=1721.344\text{ meters } \end{gathered}$

therefore, the answer is

$1721.34\text{ meters }$

I hope this helps you