- 2071 Magh · 4 marks
Velocity, acceleration at t=0.1
Short answer
From Newton forward differences with h=0.1 and p=1: dx/dt = (1/0.1)[1.49 + 0.5(-0.24) - 0.1667(-0.24) + 0.0833(0.26)] ≈ 14.32 cm/s. d²x/dt² = (1/0.1²)[-0.24 + 0 - 0.0833(0.26)] ≈ -26.17 cm/s².
If 'x' is in cm and 't' is in time then find velocity and acceleration when t = 0.1 second.
| t | 0 | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 |
|---|---|---|---|---|---|---|---|
| x | 30.13 | 31.62 | 32.87 | 33.64 | 33.95 | 33.81 | 33.24 |
Velocity = dx/dt and acceleration = d²x/dt². The time steps are equal (h = 0.1) and t = 0.1 is near the start of the table, so Newton's forward formula is used with t₀ = 0 (p = 1).
Difference table
| x | y | Δy | Δ²y | Δ³y | Δ⁴y | Δ⁵y | Δ⁶y |
|---|---|---|---|---|---|---|---|
| 0 | 30.13 | 1.49 | -0.24 | -0.24 | 0.26 | -0.27 | 0.29 |
| 0.1 | 31.62 | 1.25 | -0.48 | 0.02 | -0.01 | 0.02 | |
| 0.2 | 32.87 | 0.77 | -0.46 | 0.01 | 0.01 | ||
| 0.3 | 33.64 | 0.31 | -0.45 | 0.02 | |||
| 0.4 | 33.95 | -0.14 | -0.43 | ||||
| 0.5 | 33.81 | -0.57 | |||||
| 0.6 | 33.24 |
Take t₀ = 0, h = 0.1, so p = (0.1 − 0)/0.1 = 1.00. Differentiating Newton's forward formula with respect to t gives
dx/dt = (1/h)[Δx₀ + (2p−1)/2·Δ²x₀ + (3p²−6p+2)/6·Δ³x₀ + (4p³−18p²+22p−6)/24·Δ⁴x₀]
d²x/dt² = (1/h²)[Δ²x₀ + (p−1)Δ³x₀ + (6p²−18p+11)/12·Δ⁴x₀]
First derivative
| Term | Coefficient | Value of difference | Product |
|---|---|---|---|
| Δx₀ | 1.0000 | 1.49 | 1.4900 |
| Δ²x₀ | 0.5000 | -0.24 | -0.1200 |
| Δ³x₀ | -0.1667 | -0.24 | 0.0400 |
| Δ⁴x₀ | 0.0833 | 0.26 | 0.0217 |
Sum = 1.4317, so dx/dt = 1.4317/0.1 = 14.3167
Second derivative
| Term | Coefficient | Value of difference | Product |
|---|---|---|---|
| Δ²x₀ | 1.0000 | -0.24 | -0.2400 |
| Δ³x₀ | 0.0000 | -0.24 | 0.0000 |
| Δ⁴x₀ | -0.0833 | 0.26 | -0.0217 |
Sum = -0.2617, so d²x/dt² = -0.2617/0.1² = -26.1667
Answer: velocity at t = 0.1 s ≈ 14.32 cm/s; acceleration ≈ -26.17 cm/s² (negative: the body is slowing down).
✓ Answer checked by Claude Opus 5.5 (high effort)
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