---
title: "Velocity, acceleration at t=0.1"
subject: "Numerical Method"
chapter: 5
key: "semester_4/numerical_method:velocity-acceleration-position-time"
source: https://www.bctengineer.com/q/semester_4/numerical_method/velocity-acceleration-position-time
site: BCT Engineering
---

# Velocity, acceleration at t=0.1

Numerical Method · Chapter 5: Numerical Differentiation and Integration

- Asked in: 2071 Magh, 4 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-position-time
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-5

## Question

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 |

## 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.

## Answer

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

```text
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).

## 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 t = 0.1?
   A. p = 0.1
   B. p = 1
   C. p = 0
   D. p = 10
   Answer: B. p = (0.1 - 0)/0.1 = 1. (Key point: p value)

2. How is velocity obtained from the bracket 1.4317?
   A. Divide by h² to get 143.2
   B. Multiply by 0.1 to get 0.143
   C. Divide by h = 0.1 to get 14.32 cm/s
   D. Use it as it is
   Answer: C. dx/dt = (1/h)[…]. (Key point: Velocity step)

3. In the second-derivative formula, what happens to the Δ³ term at p = 1?
   A. Its coefficient is 1
   B. Its coefficient is 11/12
   C. It doubles
   D. Its coefficient (p-1) is zero, so it drops out
   Answer: D. The Δ³ coefficient is (p-1) = 0. (Key point: Coefficient at p = 1)

4. What does negative acceleration -26.17 cm/s² indicate?
   A. The body is slowing down
   B. The body is moving backwards
   C. The data are wrong
   D. The body has stopped
   Answer: A. Negative d²x/dt² means velocity is decreasing. (Key point: Sign)

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