A body with a mass of m = 0.3 kg begins to move uniformly and rectilinearly and in time t = 6 s

A body with a mass of m = 0.3 kg begins to move uniformly and rectilinearly and in time t = 6 s it covers a path equal to 48 m. Determine the force F acting on the body.

m = 0.3 kg.
t = 6 s.
S = 48 m.
V0 = 0 m / s.
F -?
The force F, which acts on the body, we find according to 2 Newton’s law, the product of the body’s mass m by its acceleration a: F = m * a.
The path S and the acceleration a with uniformly accelerated motion will be expressed by the formulas: S = (V ^ 2 – V0 ^ 2) / 2 * a and a = (V – V0) / t.
Since the body begins its movement from a state of rest V0, the formulas will take the form: S = V ^ 2/2 * a and a = V / t.
S = (a * t) ^ 2/2 * a = a * t ^ 2/2.
a = 2 * S / t ^ 2.
F = m * 2 * S / t ^ 2.
F = 0.3 kg * 2 * 48 m / (6 s) 2 = 0.8 N.
Answer: a force F = 0.8 N. acts on the body.