A particle of mass m is at rest at the origin at time t=0. It is subjected to a force F(t) = F0e⁻ᵇᵗ in the x direction. Its speed v(t) is depicted by which of the following curves?

F = ma = F0e⁻ᵇᵗ
Since F = ma, we have ma = F0e⁻ᵇᵗ.
Therefore, a = (F0/m)e⁻ᵇᵗ.
Acceleration is the derivative of velocity with respect to time, so we can write:
dv/dt = (F0/m)e⁻ᵇᵗ
To find the velocity v(t), we integrate both sides with respect to time:
∫dv = ∫(F0/m)e⁻ᵇᵗ dt
v(t) = -(F0/mb)e⁻ᵇᵗ + C
Since the particle is at rest at t=0, v(0) = 0. Substituting this into the equation above:
0 = -(F0/mb)e⁰ + C
C = F0/mb
Therefore, the velocity as a function of time is:
v(t) = (F0/mb)(1 - e⁻ᵇᵗ)
This equation shows that the velocity starts at 0 (when t=0) and asymptotically approaches F0/mb as time goes to infinity. The curve representing this would be an exponential growth curve that starts at 0 and levels off.