Velocity, acceleration at 0.2 s

Velocity = dy/dt and acceleration = d²y/dt². The data are equally spaced (h = 0.1), so differentiate Newton's interpolating polynomial. Since t = 0.2 is near the start of the table, use the forward formula with t₀ = 0, p = (0.2 − 0)/0.1 = 2.

Forward difference table

Formulas at p = 2

y'(t)  = (1/h)[Δy₀ + (2p−1)/2·Δ²y₀ + (3p²−6p+2)/6·Δ³y₀ + (4p³−18p²+22p−6)/24·Δ⁴y₀]
y''(t) = (1/h²)[Δ²y₀ + (p−1)Δ³y₀ + (6p²−18p+11)/12·Δ⁴y₀]

y'(0.2)  = 10[ 1.5 + 1.5(4.1) + (2/6)(-2.5) + (−2/24)(3.9) ]
         = 10 × 6.4917 = 64.9167
y''(0.2) = 100[ 4.1 + 1(-2.5) + (−1/12)(3.9) ]
         = 100 × 1.2750 = 127.5000

The coefficients at p = 2 are (2p−1)/2 = 3/2, (3p²−6p+2)/6 = 2/6, (4p³−18p²+22p−6)/24 = −2/24, (p−1) = 1 and (6p²−18p+11)/12 = −1/12.

Answer: velocity at t = 0.2 s ≈ 64.9167; acceleration ≈ 127.5000 (units of y per s and per s²).

Check with central differences (using t = 0.1, 0.2, 0.3): v ≈ (14.3 − 1.5)/0.2 = 64.0000, a ≈ (14.3 − 14.2 + 1.5)/0.01 = 160.0000. The two methods are close for velocity but the 2nd derivative differs because the data are not smooth at higher differences.