Pages

Яндекс.Метрика

Tuesday, September 20, 2016

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} / 1s9m / s
v(4.5s){x(5s) - x(4s)} / 1s{40m - 33m} / 1s7m / 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

No comments:

Post a Comment