1. Given force as a function of time

The particle starts from rest and is acted upon by $F(t)=F_0 e^{-bt}$ in the positive $x$-direction.

By Newton’s second law, $m\frac{dv}{dt}=F_0 e^{-bt}$ so the acceleration is $\frac{dv}{dt}=\frac{F_0}{m}e^{-bt}.$

2. Find the velocity function

Integrate with respect to time: $v(t)=\int \frac{F_0}{m}e^{-bt}\,dt.$

Using the initial condition $v(0)=0$, $v(t)=\frac{F_0}{m}\int_0^t e^{-b\tau}\,d\tau.$

Thus, $v(t)=\frac{F_0}{m}\left[\frac{1-e^{-bt}}{b}\right]$ so $v(t)=\frac{F_0}{mb}\left(1-e^{-bt}\right).$

3. Analyze the shape of $v(t)$

The velocity function is $v(t)=\frac{F_0}{mb}(1-e^{-bt}).$

Now observe:

• At $t=0$, $v(0)=\frac{F_0}{mb}(1-1)=0.$ So the curve starts from the origin.

• As $t\to\infty$, $e^{-bt}\to 0,$ hence $v(t)\to \frac{F_0}{mb}.$ So the speed approaches a constant limiting value.

• First derivative: $\frac{dv}{dt}=\frac{F_0}{m}e^{-bt}>0.$ Therefore $v(t)$ increases monotonically.

• Second derivative: $\frac{d^2v}{dt^2}=-\frac{bF_0}{m}e^{-bt}<0.$ Hence the graph is concave downward.

4. Conclusion about the graph

So the speed-time graph must:

• start from zero,
• rise continuously,
• have decreasing slope,
• and asymptotically approach a horizontal line.

This corresponds to option B.