Kinematics Question 358

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

Options:

A) 6 m

B) 4 m

C) $ \frac{10}{3}m $

D) $ \frac{19}{3}m $

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Answer:

Correct Answer: C

Solution:

distance travelled in the nth second is given by $ t _{n}=u+\frac{a}{2}( 2n-1 ) $

$ \text{Put u=0, a=}\frac{4}{3}m{{s}^{-2}}, $

$ n=3 $
$ \therefore d=0+\frac{4}{3\times 2}( 2\times 3-1 )=\frac{4}{6}\times 5=\frac{10}{3}m $



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