Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DMedium2 minPYQ_2024

PhysicsMediumsingle choice

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

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Answer:

Solution:

Distance travelled in the nth second is given by $t_{n} = u + \frac{a}{2} (2 n - 1)$ put $u = 0, a = \frac{4}{3} ms^{- 2}, n = 3$ $∴ d = 0 + \frac{4}{3 \times 2} (2 \times 3 - 1) = \frac{4}{6} \times 5 = \frac{10}{3} m$

Stream:BITSATSubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: PYQ_2024

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