The distance travelled by a particle starting from rest and moving with an acceleration 4/3 ms⁻², in the third second is:

Question

Accepted Answer

6m. The distance traveled in the nth second is given by the formula: $ s_n = u + \frac{1}{2}a(2n-1) $. Here, $ u = 0 $, $ a = \frac{4}{3} $, and $ n = 3 $. Substituting these values, $ s_3 = 0 + \frac{1}{2} \times \frac{4}{3} \times (2 \times 3 - 1) = 6 \, \text{m} $.

The distance travelled by a particle starting from rest and moving with an acceleration 4/3 ms⁻², in the third second is:

Solution

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