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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
Options
(a) 10/3 m
(b) 19/3 m
(c) 6 m
(d) 4 m
Correct Answer:
10/3 m
Explanation:
Distance travelled in the nth second is given by tₙ = u + a/2 (2n – 1)
put u= 0, a= 4/3 ms⁻², n= 3 d= 0 + 4/ 3×2 (2 x 3 – 1) = 4/6 x 5 = 10/ 3 m
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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