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प्रश्न
The following is the distance-time table of an object in motion:
| Time in seconds | Distance in metres |
| 0 | 0 |
| 1 | 1 |
| 2 | 8 |
| 3 | 27 |
| 4 | 64 |
| 5 | 125 |
| 6 | 216 |
| 7 | 343 |
- What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
- What do you infer about the forces acting on the object?
उत्तर
a) Here, u = 0
Using s = `ut + 1/2 at^2`
= `(at^2)/2`
∴ a = `(2s)/t^2`
| Time in Seconds | Distance in metres | `a = (2s)/t^2` |
| 0 | 0 | 0 |
| 1 | 1 | 2 |
| 2 | 8 | 4 |
| 3 | 27 | 6 |
| 4 | 64 | 8 |
| 5 | 125 | 10 |
| 6 | 216 | 12 |
| 7 | 343 | 14 |
Thus, acceleration is increasing.
b) As F = ma, therefore, F ∝ a. Hence, the force must also increase uniformly with time.
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