A particle of mass m is at rest at the origin at time t=0. It is subjected to a force F(t)=Fₒe^-bt in the x direction. Its speed v(t) is depicted by which of the following curves?

Question

Accepted Answer

(2) a curve from the origin saturating onto (Fₒ)/(mb). Idea: integrate the force to get the momentum, and read off both the starting slope and the final plateau. m(dv)/(dt)=Fₒe^-bt, so v(t)=1/m∫₀^tFₒe^-bt'dt'=(Fₒ)/(mb)(1-e^-bt). At t=0: v=0, and the slope is (Fₒ)/m, which is positive and finite — the curve leaves the origin steeply, not flat. As t→∞: e^-bt→0, so v→(Fₒ)/(mb) — a horizontal asymptote, approached from below. In between: (dv)/(dt)=(Fₒ)/me^-bt is always positive, so v never turns round. The curve therefore rises from the origin with steadily decreasing slope and levels off at (Fₒ)/(mb). A dimensional check disposes of the first option on its own: Fₒb/m has units of m s⁻³, not a speed.

A particle of mass m is at rest at the origin at time t=0. It is subjected to a force F(t)=Fₒe^-bt in the x direction. Its speed v(t) is depicted by which of the following curves?

Step-by-step solution

Why the other options are wrong