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 }`

Step-by-step explanation

to solve this we need to use the equation :

`\begin{gathered} x=v_0t+(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(m)/(s^2) \\ t=32.8\text{ sec} \\ x=x \end{gathered}`

replace

`\begin{gathered} x=v_0t+(at^2)/(2) \\ x=(o)(32.8s)+((3.2(m)/(s^2))(32.8s)^2)/(2) \\ x=0)+((3.2(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