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Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DMedium2 minQB

PhysicsMediumsingle choice

The velocity of a particle increases uniformly from $0$ to $8 m / s$ in $2$ s, stays $8 m / s$ until $t = 4$ s, then decreases uniformly, crossing zero at $t = 6$ s and reaching $- 4 m / s$ at $t = 8$ s. The distance of the particle from the origin at $t = 8$ s is:

Options:

Answer:

Solution:

Signed areas: $\frac{1}{2} (2) (8) + (2) (8) + \frac{1}{2} (2) (8) = 8 + 16 + 8 = 32$ m up to $t = 6$ s, then $\frac{1}{2} (2) (- 4) = - 4$ m, giving $32 - 4 = 28$ m. Option B.

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: QB

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