Question

An aeroplane taxies on the runway for 30 s With an acceleration of 3.2 m/s{}^{2} before taking off. how much distance would it have covered on the runway?


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Answer 1

Answer:

1.4 km

Explanation:

Time on runway = t = 30 seconds

Acceleration on the runway = a = 3.2 m/s²

Let's take the initial velocity with which the plane started as u = 0

So second equation of motion,

S=ut+1/2at²

S=0(30)+1/2×3.2×(30)²

S=1.6×900

S = 1440

Rhe distance covered by the plane is 1440 m

Or 1.4km