Question

A particle is at rest at the origin. It moves along the x -axis with an acceleration \( x - x^2 \), where x is the distance of the particle at time t. The particle next comes to rest after it has covered a distance

• A. 1
• B. \( \frac{1}{2} \)
• C. \( \frac{3}{2} \)
• D. 2

Hint:

When acceleration is given as a function of position \( x \), use the identity \( a = v \frac{dv}{dx} \). This allows you to set up an integrable equation \( \int v \, dv = \int a(x) \, dx \) to find the velocity \( v \) as a function of \( x \).

Answer 1

The Correct Option is C

Let the acceleration be \( a \) and velocity be \( v \). We are given \( a = x - x^2 \). We know that acceleration can be written as \( a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx} \). So, we have the differential equation: \[ v \frac{dv}{dx} = x - x^2 \] Separate the variables and integrate: \[ \int v \, dv = \int (x - x^2) \, dx \] \[ \frac{v^2}{2} = \frac{x^2}{2} - \frac{x^3}{3} + C \] The particle starts from rest at the origin, which means when \( x = 0 \), \( v = 0 \). Substituting these initial conditions to find C: \[ \frac{0^2}{2} = \frac{0^2}{2} - \frac{0^3}{3} + C \Rightarrow C = 0 \] So, the relationship between velocity and position is: \[ \frac{v^2}{2} = \frac{x^2}{2} - \frac{x^3}{3} \] The particle next comes to rest when \( v = 0 \) again, for some \( x > 0 \). \[ 0 = \frac{x^2}{2} - \frac{x^3}{3} \] \[ \frac{x^3}{3} = \frac{x^2}{2} \] Since we are looking for a distance \( x > 0 \), we can divide by \( x^2 \): \[ \frac{x}{3} = \frac{1}{2} \] \[ x = \frac{3}{2} \] The particle next comes to rest after covering a distance of \( \frac{3}{2} \).