The formula to calculate the Distance Traveled (D) in the Nth Second of Accelerated Translatory Motion is:
\[ D = u + \left(\frac{2n - 1}{2}\right) \cdot a \]
Where:
Distance Traveled defines how much path an object has covered to reach its destination in a given period.
Initial Velocity is the velocity at which motion starts.
The Nth Second is the n seconds time covered by the body.
Acceleration of Body is the rate of change in velocity to the change in time.
Let's assume the following values:
Using the formula:
\[ D = 35 + \left(\frac{2 \cdot 4 - 1}{2}\right) \cdot 4.8 \]
Evaluating:
\[ D = 35 + \left(\frac{7}{2}\right) \cdot 4.8 = 35 + 3.5 \cdot 4.8 = 35 + 16.8 = 51.8 \]
The Distance Traveled is 51.8 meters.
| Initial Velocity (u) (m/s) | Nth Second (n) (s) | Acceleration (a) (m/s²) | Distance Traveled (D) (m) |
|---|---|---|---|
| 30 | 2 | 4 | 36.0000 |
| 30 | 2 | 4.8 | 37.2000 |
| 30 | 2 | 5.6 | 38.4000 |
| 30 | 4 | 4 | 44.0000 |
| 30 | 4 | 4.8 | 46.8000 |
| 30 | 4 | 5.6 | 49.6000 |
| 30 | 6 | 4 | 52.0000 |
| 30 | 6 | 4.8 | 56.4000 |
| 30 | 6 | 5.6 | 60.8000 |
| 35 | 2 | 4 | 41.0000 |
| 35 | 2 | 4.8 | 42.2000 |
| 35 | 2 | 5.6 | 43.4000 |
| 35 | 4 | 4 | 49.0000 |
| 35 | 4 | 4.8 | 51.8000 |
| 35 | 4 | 5.6 | 54.6000 |
| 35 | 6 | 4 | 57.0000 |
| 35 | 6 | 4.8 | 61.4000 |
| 35 | 6 | 5.6 | 65.8000 |
| 40 | 2 | 4 | 46.0000 |
| 40 | 2 | 4.8 | 47.2000 |
| 40 | 2 | 5.6 | 48.4000 |
| 40 | 4 | 4 | 54.0000 |
| 40 | 4 | 4.8 | 56.8000 |
| 40 | 4 | 5.6 | 59.6000 |
| 40 | 6 | 4 | 62.0000 |
| 40 | 6 | 4.8 | 66.4000 |
| 40 | 6 | 5.6 | 70.8000 |