---
title: "Velocity, acceleration at 0.2 s"
subject: "Numerical Method"
chapter: 5
key: "semester_4/numerical_method:velocity-acceleration-distance-0-2"
source: https://www.bctengineer.com/q/semester_4/numerical_method/velocity-acceleration-distance-0-2
site: BCT Engineering
---

# Velocity, acceleration at 0.2 s

Numerical Method · Chapter 5: Numerical Differentiation and Integration

- Asked in: 2069 Poush, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/velocity-acceleration-distance-0-2
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-5

## Question

The distance y(t) traversed in time t by an object moving in a straight line is given below; approximate the velocity and acceleration at 0.2 seconds.
| t (in seconds) | 0.0 | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 |
|---|---|---|---|---|---|---|---|
| y (in times) | 0.0 | 1.5 | 7.1 | 14.3 | 24.5 | 36.7 | 50.0 |

## Short answer

Use Newton forward differences with h = 0.1 and p = 2. The velocity at t = 0.2 s is approximately 64.9167 and the acceleration approximately 127.5000 in the given units.

## Answer

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
| t | y | Δy | Δ²y | Δ³y | Δ⁴y |
|---|---|---|---|---|---|
| 0 | 0.0 | 1.5 | 4.1 | -2.5 | 3.9 |
| 0.1 | 1.5 | 5.6 | 1.6 | 1.4 | -2.4 |
| 0.2 | 7.1 | 7.2 | 3.0 | -1.0 | 0.1 |
| 0.3 | 14.3 | 10.2 | 2.0 | -0.9 |  |
| 0.4 | 24.5 | 12.2 | 1.1 |  |  |
| 0.5 | 36.7 | 13.3 |  |  |  |
| 0.6 | 50.0 |  |  |  |  |

#### Formulas at p = 2
```text
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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With t0 = 0, h = 0.1, what is p at 0.2?
   A. p = 2
   B. p = 0.2
   C. p = 1
   D. p = 0.5
   Answer: A. p = 0.2/0.1. (Key point: p)

2. What is (2p-1)/2 at p = 2?
   A. 1/2
   B. 3/2
   C. 3
   D. 2
   Answer: B. (4-1)/2 = 1.5. (Key point: Coefficient of Δ²)

3. What velocity results?
   A. 6.49167
   B. 649.167
   C. 64.9167
   D. 127.5
   Answer: C. y' = (1/h)(6.4917) = 10 × 6.4917. (Key point: Velocity)

4. Why is the bracket multiplied by 100?
   A. h = 0.01 is used
   B. To convert units
   C. Because p = 2 squared is 4
   D. The second derivative has 1/h² = 1/0.01
   Answer: D. y'' = (1/h²)[…] = 127.5. (Key point: Acceleration scale)

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