Question

an aeroplane on a runway of 90 s with 6.4 m/s of accelaration before takeoff how much distance would it covered on runway

Answer 1

$s = \frac{1}{2}a {t}^{2}$
$s = \frac{1}{2} \times 6.4 \times {90}^{2}$
$s = 3.2 \times 8100 = 25920 \: meters$
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Answer 2

Answer:

The aeroplane would have covered 25920 m distance.

Explanation:

a = 6.4 m/s²

t = 90 s

u = 0 m/s

s = ut + ½ at²

=> s = 0 * 90 + ½ * 6.4 * 90²

=> s = 0 + ½ * 6.4 * 8100

=> s = ½ * 8100 * 6.4

=> s = 4050 * 6.4

=> s = 25920 m

The aeroplane would have covered 25920 m distance.