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Physics - Kinematics Question with Solution | TestHub

PhysicsKinematicsMiscellaneousHard2 minQB

PhysicsHardmultiple choice

A particle is moving in x-y plane and equation of its path is given by $y = x^{3} + 2 x^{2} + 5 x + 3$ . At $t = 0$ particle is located at $x = 0$ . Consider the case for $x \geq 0, y \geq 0$ . Choose the CORRECT option(s):-

Options:(select one or more)

Answer:

A, B, D

Solution:

$tan θ = \frac{dy}{dx} = 3 x^{2} + 4 x + 5$

(A) At $t = 0$ , $x = 0$ , $tan θ = 5$ .

(D) At x = -4/3, velocity of the particle makes the same angle as it does at the point of projection, which is not included in the given case.

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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