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A plane accelerates from rest at a constant rate of $5.00 \, \frac{m}{s^2}$ along a runway that is $1800\;m$ long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time $t_{TO}$ needed to take off?

I tried to find time by using only distance and acceleration, so $\frac{v_f-v_i}{a} = \frac{1800}{5} = 360 \, s$. However, this is incorrect. I also tried taking the square root of the answer, however that too is incorrect. Any help would be appreciated.

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The length of the runway is not the final velocity... –  User58220 Sep 25 '13 at 6:02
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1 Answer 1

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The acceleration is $a=5$, and you start from rest at position zero on the runway. The runway length is $L = 1800$.

Your speed will be $v(t) = \int_0^t a d \tau = at$. (Starting from rest means $v(0) = 0$.)

Your position will be $x(t) = \int_0^t v(\tau) d \tau = \int_0^t a \tau d \tau = a \frac{t^2}{2}$. (Starting from 'position zero' means we are taking $x(0) = 0$.)

So we need to solve $L = x(t) = a \frac{t^2}{2}$ for $t$.

This gives $t = \sqrt{\frac{2L}{a}} = \sqrt{720}$ ($\approx 27$ seconds).

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Thank you! How can i calculate the speed at vT0 when the plane takes off? What is the equation for it? –  Lisa Paima Sep 25 '13 at 6:14
    
Already figured it out. Never mind, thankyou –  Lisa Paima Sep 25 '13 at 6:18
    
$v(t) = a t = a \sqrt{\frac{2L}{a}} = \sqrt{2La}$ (appropriate if you are flying to Los Angeles). –  copper.hat Sep 25 '13 at 6:20
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