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The distance travelled by a particle starting from rest and moving with an acceleration 4/3 ms–2, in the third second is:
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- 6 m
- 4 m
- 10/3 m
- 19/3 m
Correct Option: C
Distance travelled in the nth second is given by tn = u + | (2n - 1) | |
2 |
put u = 0, 4/3 ms-2 , n = 3
∴ d = 0 + | (2 × 3 –1) = | × 5 = | m | |||
3 × 2 | 6 | 3 |