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