A particle moves along the x-axis according to x = ut e⁻ᵏᵗ, where u, k>0. After its first reversal of direction, the particle again gains speed and subsequently slows down. The maximum speed attained after this reversal and the corresponding position are:
Options

Question

A particle moves along the x-axis according to x = ut e⁻ᵏᵗ, where u, k>0. After its first reversal of direction, the particle again gains speed and subsequently slows down. The maximum speed attained after this reversal and the corresponding position are:

Options

G1 Physics · Accepted Answer

Differentiate: v = dx/dt = u e⁻ᵏᵗ(1−kt). The first reversal occurs when v=0: 1−kt=0 ⇒ t=1/k. After this, v<0, so speed is |v| = u e⁻ᵏᵗ(kt−1). Let y=kt. Then |v| = u e⁻ʸ(y−1). For maximum speed, d/dy [e⁻ʸ(y−1)] = e⁻ʸ(2−y)=0 ⇒ y=2 ⇒ t=2/k. Thus |v|ₘₐₓ = u e⁻² = u/e². Position at this instant: x = u(2/k)e⁻² = 2u/(ke²). Therefore option B is correct.

A particle moves along the x-axis according to x = ut e⁻ᵏᵗ, where u,k>0. After its first reversal of direction, the particle again gains speed and subsequently slows down. The maximum speed attained after this reversal and the corresponding position are: