A particle is projected with velocity 𝑣0 along 𝑥 − 𝑎𝑥𝑖𝑠.The deceleration on the particle is proportional to the square of the distance from the origin i.e., 𝑎 = −𝑎𝑥 2 .The distance at which the particle stops is

Question : A particle is projected with velocity 𝑣0 along 𝑥 − 𝑎𝑥𝑖𝑠.The deceleration on the particle is proportional to the square of the distance from the origin i.e., 𝑎 = −𝑎𝑥 2 .The distance at which the particle stops is

(A) 3\sqrt{\dfrac{v_{0}}{2\alpha }}

(B) \left( \dfrac{3v_{0}}{2\alpha }\right) ^{\dfrac{1}{3}}

(C) \sqrt{\dfrac{3v_{0}^{2}}{2\alpha \cdot }}

(D) \left( \dfrac{3v_{0}^{2}}{2\alpha }\right) ^{\dfrac{1}{3}}

Answer : option (D)

Solution :

A particle is projected with velocity 𝑣0 along 𝑥 − 𝑎𝑥𝑖𝑠.The deceleration on the particle is proportional to the square of the distance from the origin i.e., 𝑎 = −𝑎𝑥 2 .The distance at which the particle stops is
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