June 3rd, 2017, 04:16 AM  #1 
Newbie Joined: Jun 2017 From: Bolton Posts: 2 Thanks: 0  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 takeoff velocity. Calculate the takeoff 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 
June 3rd, 2017, 04:41 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,770 Thanks: 1424 
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 ... $v_f=v_0+at \implies t = \dfrac{v_fv_0}{a}$ $\Delta x = \bar{v} \cdot t = \left(\dfrac{v_f+v_0}{2}\right) \cdot t$ substitute $\dfrac{v_fv_0}{a}$ for time in the $\Delta x$ equation ... $\Delta x= \left(\dfrac{v_f+v_0}{2}\right) \cdot \left(\dfrac{v_fv_0}{a}\right)$ $\Delta x= \dfrac{v_f^2v_0^2}{2a}$ solving for $v_f^2$ yields the "no time" kinematics equation cited above. 
June 3rd, 2017, 11:12 AM  #3 
Newbie Joined: Jun 2017 From: Bolton Posts: 2 Thanks: 0 
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 
June 3rd, 2017, 12:29 PM  #4  
Math Team Joined: Jul 2011 From: Texas Posts: 2,770 Thanks: 1424  Quote:
distance to reach takeoff 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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