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
A.
$$6\,m$$
B.
$$4\,m$$
C.
$$\frac{{10}}{3}m$$
D.
$$\frac{{19}}{3}m$$
Answer :
$$\frac{{10}}{3}m$$
Solution :
Distance travelled in $${n^{th}}$$ second is given by
$${s_n} = u + \frac{1}{2}a\left( {2n - 1} \right)$$
Here, $$u = 0,\,a = \frac{4}{3}$$
$$\therefore {s_3} = 0 + \frac{1}{2} \times \frac{4}{3} \times \left( {6 - 1} \right) = \frac{{10}}{3}m$$
Releted MCQ Question on Basic Physics >> Kinematics
Releted Question 1
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