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# An airplane needs a 300 m length of runway to reach take-off speed of 30 m/second.. What is the acceleration?

What is the formula?

### 4 Answers

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• (Vfin^2-Vin^2) = 2*a*d

acceleration a = (30^2-0)/(2*300) = 900/600 = 9/6 = 3/2 =  1.50 m/sec^2

• Law of uniformly accelerated motion:

S = 1/2 a t^2;

v = a * t;

v = 30 m/s;

S = 300 m.

a = v / t;

a = 30 / t;

1/2 * (30/t) * t^2 = 300;

15 * t = 300;

t = 300 / 15 = 20 s; (time).

a = 30 / 20 = 1.5 m/s^2; (acceleration).

Ciao.

• I use distance = (1/2) a * t^2

And V[final] = a * t

• If the acceleration is constant then v^2 = 2as -> a = v^2/ (2s) = 30*30/(2*300) = 1.5 m/s^2

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