An aeroplane has a maximum acceleration of 6.0 m s–2. Calculate the minimum length for a straight runway so that the aeroplane, starting from rest, can safely take off at a velocity of 90 m s−1.
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s=v2−v022a=902−022⋅6=675 (m)s=\frac{v^2-v_0^2}{2a}=\frac{90^2-0^2}{2\cdot 6}=675\ (m)s=2av2−v02=2⋅6902−02=675 (m) . Answer
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