Question

A particle projected with initial velocity U along X axis such that it suffers deaccelaration given by Ax^3(where A is constant).the distdistance X at which the particle is...

Answer 1

Answer on Question #48671, Physics, Mechanics | Kinematics | Dynamics

A particle projected with initial velocity $U$ along $X$ axis such that it suffers deceleration given by $Ax^3$ (where $A$ is constant). The distance $X$ at which the particle is...

Solution:

$$a(x) = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$$

Thus, integrating

$$a(x)dx = vdv$$ $$\int_{0}^{x} -Ax^3 dx = \int_{u}^{0} vdv$$ $$-A \frac{x^4}{4} = -\frac{u^2}{2}$$ $$x^4 = 2Au^2$$ $$x = (2Au^2)^{\frac{1}{4}}$$

Answer: $x = (2Au^2)^{\frac{1}{4}}$

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