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Physics - Kinematics Question with Solution | TestHub

PhysicsKinematics1-DHard2 minQB

PhysicsHardsingle choice

A car can accelerate at a maximum of $4 m / s^{2}$ and brake at a maximum retardation of $8 m / s^{2}$ . The minimum time in which it can cover $1875$ m, starting and ending at rest, is:

Options:

Answer:

Solution:

With peak speed $v_{m}$ : distance = $\frac{v _{m}^{2}}{2 ( 4 )} + \frac{v _{m}^{2}}{2 ( 8 )} = \frac{3 v _{m}^{2}}{16} = 1875$ , so $v_{m} = 100$ m/s. Then $t = \frac{100}{4} + \frac{100}{8} = 25 + 12.5 = 37.5$ s. Option C.

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:1-D

⏱ 2mℹ️ Source: QB

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