Question

A particle initially at rest, starts moving in straight
line with uniform acceleration 4/3 m/s square the distance covered in third second of its motion is
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Answer 1

Answer:46euhdhejhejhhggfdd

Explanation:

Answer 2

Answer:

The distance covered in third second of its motion is 3.33 m

Explanation:

Initial velocity of the particle is zero and the acceleration of the particle is given as $4 / 3 \mathrm{m} / \mathrm{s}^{2}$. Applying Newton law of motion. Velocity at end of 2 second will be

V = u +at

$V=0+\frac{4}{3} \times 2=\frac{8}{3} \mathrm{m} / \mathrm{s}$

Distance travelled in third second will be given by newton law again

$S=u t+1 / 2 a t^{2}$

Since u equal V in this case  

$\begin{array}{l}{\mathrm{S}=\frac{8}{3} \mathrm{x} 1+1 / 2 \frac{4}{3}} \\ {\mathrm{S}=\frac{8}{3}+\frac{2}{3}=10 / 3=3.33 \mathrm{m}}\end{array}$