A particle is projected with velocity v0 along x-axis. The retardation of the particle is proportional to the displacement from origin, a=–αx. The distance at which particle stops first time is
Question
A particle is projected with velocity v0 along x-axis. The retardation of the particle is proportional to the displacement from origin, a=–αx. The distance at which particle stops first time is
Solution
The given equation is a = -αx, which is a differential equation.
Step 1: Rewrite the equation in terms of velocity and displacement.
The acceleration a is the derivative of velocity v with respect to time t, and velocity v is the derivative of displacement x with respect to time t. So, we can write the equation as dv/dt = -αx.
Step 2: Separate the variables.
We can rewrite the equation as dv = -αx dt.
Step 3: Integrate both sides.
The integral of dv from v0 to 0 (since the particle stops, the final velocity is 0) is equal to the integral of -αx dt from 0 to x (the distance at which the particle stops).
∫v0 to 0 dv = -α ∫0 to x x dt
This gives -v0 = -α/2 * x^2.
Step 4: Solve for x.
x = sqrt((2*v0)/α)
So, the distance at which the particle stops for the first time is sqrt((2*v0)/α).
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