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Question

Physics Question on Acceleration

The distance travelled by a particle starting from rest and moving with an acceleration 43ms2,\frac{4}{3}ms^{-2}, in the third second is

A

103m\frac{10}{3}m

B

193m\frac{19}{3}m

C

6m6\, m

D

4m4\, m

Answer

103m\frac{10}{3}m

Explanation

Solution

Distance travelled in the 3rd3^{rd} second
= Distance travelled in 3s3\,s
- distance travelled in 2 s.
As, u = 0,
S(3rds)=12a.3212a.22=12.a.5S_{(3^{rd}s)}=\frac{1}{2}a.3^2 -\frac{1}{2}a.2^2=\frac{1}{2}.a.5
Given a=43ms2;S(3rds)=12×43×5=103ma=\frac{4}{3}ms^{-2}; \therefore \, \, \, \, S_{(3^{rd}s)} =\frac{1}{2} \times \frac{4}{3} \times 5 =\frac{10}{3}m