A particle moves along the x-axis such that its velocity v varies with position x as v = alpha * sqrt(x), where alpha is a constant. If the particle starts from rest at x = 0 at t = 0, what is the acceleration of the particle at any time t?

A. ^22 — Using the relation a = v * dv/dx, we find the acceleration is constant. Since v = alpha * x^(1/2), dv/dx = alpha / (2 * x^(1/2)), thus a = (alpha * x^(1/2)) * (alpha / (2 * x^(1/2))) = alpha^2 / 2.

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$A particle moves along the x-axis such that its velocity v varies with position x as v = α x, where α is a constant. If the particle starts from rest at x = 0 at t = 0, what is the acceleration of the particle at any time t ?$

A. $\frac{α ^{2}}{2}$
B. $\frac{α ^{2} t}{2}$
C. $\frac{α ^{2} t ^{2}}{4}$
D. $α^{2}$

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Correct answer: A — $\frac{α ^{2}}{2}$

Solution

Using the relation a = v \frac{d v}{d x}, we find the acceleration is constant. Since v = α x^{1/2}, \frac{d v}{d x} = \frac{α}{2 x ^{1/2}}, thus a = (α x^{1/2}) \cdot (\frac{α}{2 x ^{1/2}}) = \frac{α ^{2}}{2} .

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A particle moves along the x-axis such that its velocity v varies with position x as v=αx​, where α is a constant. If the particle starts from rest at x=0 at t=0, what is the acceleration of the particle at any time t?

Solution

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