An aeroplane lands at 216km/h and stops after covering a runway of 2 km. Calculate the acceleration and the time in which it's comes to rest

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Solution

Dear student

$u \sin g t h e f o r m u l a v^{2} - u^{2} = 2 \times a \times x 0^{2} - 216^{2} = 2 \times a \times 2 a = \frac{- 216 \times 216}{4} = - 11664 k m / h r^{2} u \sin g t h e f o r m u l a v = u + a t 0 = 216 - \frac{216 \times 216}{4} t 216 = \frac{216 \times 216}{4} t 1 = \frac{216}{4} t t = \frac{4}{216} = 0.018 h r R e g a r d s$

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An aeroplane lands at 216km/h and stops after covering a runway of 2 km. Calculate the acceleration and the time in which it's comes to rest

Dear student using the formula v2-u2=2×a×x02-2162=2×a×2a=-216×2164=-11664km/hr2using the formula v=u+at0=216-216×2164t216=216×2164t1=2164tt=4216=0.018hrRegards