- 2072 Asoj · 6 marks
Velocity and acceleration at t=3
Short answer
Using Newton forward differences with h=2 min, p=0.5. Velocity = (1/h)[Δs0 + (2p-1)/2 Δ²s0 + …] = 0.3708 km/min (≈ 22.25 km/h). Acceleration = (1/h²)[Δ²s0 + (p-1)Δ³s0 + …] = 0.0901 km/min².
The distance travelled by a vehicle at intervals of 2 minutes are given as follows:
| Time (min) | 2 | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|---|
| Distance (km) | 0.25 | 1 | 2.2 | 4 | 6.5 | 8.5 |
| Evaluate the velocity and acceleration of the vehicle at t = 3 minutes. |
Let s be the distance (km) and t the time (min). Velocity = ds/dt and acceleration = d²s/dt². The steps are equal (h = 2), and t = 3 lies near the start of the table, so Newton's forward formula is used.
Difference table
| x | y | Δy | Δ²y | Δ³y | Δ⁴y | Δ⁵y |
|---|---|---|---|---|---|---|
| 2 | 0.25 | 0.75 | 0.45 | 0.15 | -0.05 | -1.25 |
| 4 | 1.00 | 1.20 | 0.60 | 0.10 | -1.30 | |
| 6 | 2.20 | 1.80 | 0.70 | -1.20 | ||
| 8 | 4.00 | 2.50 | -0.50 | |||
| 10 | 6.50 | 2.00 | ||||
| 12 | 8.50 |
Take t₀ = 2, h = 2, so p = (3 − 2)/2 = 0.50. Differentiating Newton's forward formula with respect to t gives
ds/dt = (1/h)[Δs₀ + (2p−1)/2·Δ²s₀ + (3p²−6p+2)/6·Δ³s₀ + (4p³−18p²+22p−6)/24·Δ⁴s₀]
d²s/dt² = (1/h²)[Δ²s₀ + (p−1)Δ³s₀ + (6p²−18p+11)/12·Δ⁴s₀]
First derivative
| Term | Coefficient | Value of difference | Product |
|---|---|---|---|
| Δs₀ | 1.0000 | 0.75 | 0.7500 |
| Δ²s₀ | 0.0000 | 0.45 | 0.0000 |
| Δ³s₀ | -0.0417 | 0.15 | -0.0062 |
| Δ⁴s₀ | 0.0417 | -0.05 | -0.0021 |
Sum = 0.7417, so ds/dt = 0.7417/2 = 0.3708
Second derivative
| Term | Coefficient | Value of difference | Product |
|---|---|---|---|
| Δ²s₀ | 1.0000 | 0.45 | 0.4500 |
| Δ³s₀ | -0.5000 | 0.15 | -0.0750 |
| Δ⁴s₀ | 0.2917 | -0.05 | -0.0146 |
Sum = 0.3604, so d²s/dt² = 0.3604/2² = 0.0901
Answer: velocity at t = 3 min ≈ 0.3708 km/min (≈ 22.25 km/h); acceleration ≈ 0.0901 km/min²
The Δ⁵ term (Δ⁵s₀ = −1.25) is left out above because the higher differences are irregular. If it is included (the full degree-5 polynomial), the velocity is ≈ 0.394 km/min and the acceleration ≈ 0.149 km/min².
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- Velocity, acceleration at 0.2 s · Numerical Method, chapter 5
- Velocity, acceleration at t=0.1 · Numerical Method, chapter 5
- Angular velocity from rod data · Numerical Method, chapter 5 · asked 2×
- Acceleration proportional to time; distance travelled · Applied Mechanics, chapter 8
- Maximum speed and distance before braking (2.5 km) · Utilization of Electrical Energy, chapter 4
- Sewer diameter for a 12 sq km area · Sanitary Engineering, chapter 3
- Velocity and acceleration of a particle · Mathematics III, chapter 2