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June 3rd, 2017, 04:16 AM   #1
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Initial Velocity, Final Velocity

I'm very stuck on this question:

A small aeroplane used for skydiving moves along a runway.
The aeroplane accelerates at 2 m / s^2 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.

The mark scheme says the answer is

v^2 = (2 x 2 x 209)+ 8^2
v = 30 m / s

but I am unsure of how to arrive at that answer. If someone could explain this question to me, that would be much appreciated. Thank you
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June 3rd, 2017, 04:41 AM   #2
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The equation, $v_f^2 = v_0^2+2a \Delta x$, is one of the basic kinematics equations for motion with uniform acceleration.
Note the parameter of time has been eliminated.

It is derived as follows ...

$v_f=v_0+at \implies t = \dfrac{v_f-v_0}{a}$

$\Delta x = \bar{v} \cdot t = \left(\dfrac{v_f+v_0}{2}\right) \cdot t$

substitute $\dfrac{v_f-v_0}{a}$ for time in the $\Delta x$ equation ...

$\Delta x= \left(\dfrac{v_f+v_0}{2}\right) \cdot \left(\dfrac{v_f-v_0}{a}\right)$

$\Delta x= \dfrac{v_f^2-v_0^2}{2a}$

solving for $v_f^2$ yields the "no time" kinematics equation cited above.
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June 3rd, 2017, 11:12 AM   #3
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Thanks, but I just want to know where the values come from...
And if your answer already tells me that, then I cannot understand your answer
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June 3rd, 2017, 12:29 PM   #4
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Quote:
A small aeroplane used for skydiving moves along a runway.
The aeroplane accelerates at 2 m / s^2 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.
The question is asking for take-off velocity ... that would be $v_f$

distance to reach take-off velocity = 209 m ... that would be $\Delta x$

airplane accelerates at 2m/s^2 ... that would be acceleration, $a$

from a velocity of 8 m/s ... that would be $v_0$

$v_f^2 = v_0^2 + 2a \Delta x$

$v_f^2 = 8^2 + 2(2)(209)$

$v_f = \sqrt{8^2 + 2(2)(209)} = \sqrt{900} = 30 \, m/s$

clear?
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