With what acceleration did a jet plane weighing 60 tons move during takeoff, if the thrust force of the engine was 90 kN?

In the vertical direction, two forces act on a jet plane: gravity Ft from the Earth and the reaction force of the support N (elasticity of the runway). Both forces compensate for each other’s action, they are oppositely directed and equal in magnitude: N = Fт = m ∙ g, where the proportionality coefficient g = 9.8 N / kg. In the horizontal direction, the thrust force of the engine Fдв and the friction force Fтр = μ ∙ N = μ ∙ m ∙ g act on a jet plane, where μ is the coefficient of friction. In this case, the plane receives a horizontal acceleration a. According to Newton’s second law, F = m ∙ a, that is, the resultant will be: F = Fdv – Ftr. It turns out that during the takeoff run, a jet plane with a mass of m = 60 t = 60,000 kg, and an engine thrust force of Fm = 90 kN = 90,000 N, will receive an acceleration a = F: m, a = (Fdw – Ftr): m. Since the problem does not talk about the coefficient of friction μ, it means that the friction force can be neglected, then a = 90,000 N: 60,000 kg = 1.5 m / s ^ 2.
Answer: a = 1.5 m / s ^ 2.