Question

c) A small aeroplane used for skydiving moves along a runway. The aeroplane accelerates at 2 m/s² from a velocity of 8 m/s. After a distance of 209 m it reaches its take-off velocity. Calculate the take-off velocity of the aeroplane.​

Answer 1

`30\; {\rm m\cdot s^(-1)}`.

Step-by-step explanation:

Since acceleration stays the same, Apply the SUVAT equation
`v^(2) - u^(2) = 2\, a\, x`, where:

• 
  `v` is the velocity of the airplane after accelerating, which needs to be found;
• 
  `u = 8\; {\rm m\cdot s^(-1)}` is the velocity of the airplane before accelerating;
• 
  `a = 2\; {\rm m\cdot s^(-2)}` is the acceleration of the airplane;
• 
  `x = 209\; {\rm m}` is the displacement of the airplane during the acceleration.

Rearrange this equation to find
`v`, velocity of the plane after accelerating:

`v^(2) = u^(2) + 2\, a\, x`.

`\begin{aligned} v &= \sqrt{u^(2) + 2\, a\, x} \\ &= \sqrt{(8)^(2) + 2\, (2)\, (209)}\; {\rm m\cdot s^(-1)} \\ &= √(900)\; {\rm m\cdot s^(-1)} \\ &= 30\; {\rm m\cdot s^(-1)}\end{aligned}`.