I've found the complete form of this problem from another website which is shown in the attached picture. The equation is:
F(t) = -25 + (1.23×10⁵)t - (5.58×10⁶)t²
a.) For the first question, let's substitute t = 22.4×10⁻³ seconds to the formula.
F = -25 + (1.23×10⁵)(22.4×10⁻³ s) - (5.58×10⁶)(22.4×10⁻³ s)² F = -69.62 N From Newton's second law, F = ma. -69.62 = (165 g)(1 kg/1000 g)(a) Solving for a, a = -421.94 m/s² = Δv/Δt = (0 - v)/(22.4×10⁻³ - 0) Solving for v, v = 9.45 m/s
b.) To solve for the distance, the formula is:
d = v₀t + 1/2(a)(t²) Let's use the absolute value of a because distance is always positive. d = 0(22.4×10⁻³) + 1/2(421.94)(22.4×10⁻³)² d = 0.106 m
