Question

An aircraft starts at rest and is accelerated for 8.6 seconds, at which time the aircraft launches. If the distance traveled from the starting point to the launch point was 522.6 m, what was the launch velocity of the aircraft in m/s

Answer 1

Answer:121.53 m/s

Step-by-step explanation:

Given

Time taken
`t=8.6\ s`

Distance traveled
`s=522.6\ m`

initial velocity
`u=0`

using
`s=ut+(1)/(2)at^2`

`522.6=0+(1)/(2)* a* (8.6)^2`

`a=14.131\ m/s^2`

To find the launch velocity

`v=u+at`

`v=0+(14.131)\cdot 8.6`

`v=121.53\ m/s`