Question:

The distance travelled by a particle starting from rest and moving with an acceleration \( \frac{4}{3} \) ms\(^{-2} \), in the third second is:

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The displacement in the \( n \)th second formula: \[ s_n = u + \frac{a}{2} (2n - 1) \] is useful for determining the exact distance covered in a given second without calculating total displacement.
Updated On: Mar 24, 2025
  • \( 6 \) m
  • \( 4 \) m
  • \( \frac{10}{3} \) m
  • \( \frac{19}{3} \) m
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The Correct Option is C

Solution and Explanation

Step 1: {Use the nth second displacement formula} 
The displacement covered in the \( n \)th second is given by: \[ s_n = u + \frac{a}{2} (2n - 1) \] where: - \( u \) is the initial velocity, - \( a \) is the acceleration, - \( n \) is the time instant. 
Step 2: {Substituting values} 
Given: \[ u = 0, \quad a = \frac{4}{3} { ms}^{-2}, \quad n = 3 \] \[ s_3 = 0 + \frac{\frac{4}{3}}{2} (2(3) - 1) \] \[ = \frac{4}{6} \times 5 = \frac{10}{3} \,m \] Step 3: {Verify the options} 
Thus, the correct answer is (C) \( \frac{10}{3} \) m. 
 

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