Problem: The figure shows the position of a car (black circles) at one-second intervals. What is the acceleration at the time t = 4 s

Problem

The figure shows the position of a car (black circles) at one-second intervals.
What is the acceleration at the time t = 4 s?
 

Solution 1:

x(3s) = 24m

x(4s) = 33m

x(5s) = 40m

v(3.5s) ≈ {x(4s) - x(3s)} / 1s ≈ {3m - 24m} / 1s ≈ 9m / s

v(4.5s) ≈ {x(5s) - x(4s)} / 1s ≈ {40m - 33m} / 1s ≈ 7m / s

a(4s) ≈ {v(4.5s) - v(3.5s)} / 1s ≈ {7m / s - 9m / s} / 1s ≈ { - 2m / s} / 1s ≈ - 2m / s²

Solution 2:

a(t) ≈ {v(t + ½Δt) - v(t - ½Δt)} / Δt

v(t) ≈ {x(t + ½Δt) - x(t - ½Δt)} / Δt

v(t + ½Δt) ≈ {x(t+Δt) - x(t)} / Δt

v(t - ½Δt) ≈ {x(t) - x(t-Δt )} / Δt

a(t) ≈ ([{x(t+Δt) - x(t)} / Δt] - [{x(t) - x(t-Δt )} / Δt]) / Δt

a(t) ≈ ({x(t+Δt) - x(t)} - {x(t) - x(t-Δt )}) / (Δt)²

a(t) ≈ {x(t+Δt) - 2x(t) + x(t-Δt )} / (Δt)² derived important formula

Δt = 1s

t = 4s

a(4s) ≈ {x(5s) - 2x(4s) + x(3s)} / (1s)²

a(4s) ≈ {40m - 2 ∙ 33m + 24m} / 1s² ≈ {40m - 66m + 24m} / 1s² ≈ - 2m / 1s² ≈ - 2m / s²

Solution 3:

if a = const

x(t+Δt) = x(t) + v(t) ∙ Δt + ½ ∙ a ∙ (Δt)²

x(t-Δt) = x(t) - v(t) ∙ Δt + ½ ∙ a ∙ (Δt)²

x(t+Δt) + x(t-Δt) = [x(t) + v(t) ∙ Δt + ½ ∙ a ∙ (Δt)²] + [x(t) - v(t) ∙ Δt + ½ ∙ a ∙ (Δt)²]

x(t+Δt) + x(t-Δt) = 2x(t) + a ∙ (Δt)²

x(t+Δt) + x(t-Δt) - 2x(t) = a ∙ (Δt)²

x(t+Δt) - 2x(t) + x(t-Δt) = a ∙ (Δt)²

a = {x(t+Δt) - 2x(t) + x(t-Δt)} / (Δt)² derived important formula

Δt = 1s

t = 4s
a = {x(5s) - 2x(4s) + x(3s)} / (1s)²

a = {40m - 2 ∙ 33m + 24m} / 1s² = {40m - 66m + 24m} / 1s² = - 2m / 1s² = - 2m / s²