Animated Solution for Physics - Laws of Motion: A particle of mass $m$ is at rest at the origin at time $t=0$. It is subjected to a force $F(t)=F_0{e}^{-bt}$ in the x-direction. Its speed $v(t)$ is depicted by which of the following curves?

Unraveling the Velocity-Time Graph

Imagine a particle sitting perfectly still at the origin. Suddenly, a force kicks in. But this isn't your everyday constant push. This force is given by the expression $F(t)=F_0{e}^{-bt}$.

What does that mean physically? It means the force is strongest at the very beginning, right at time $t=0$. As time ticks on, this force exponentially fades away, getting weaker and weaker. Our mission is to figure out how the speed of this particle changes over time and to select the correct velocity-time graph from the given options.

To understand the velocity, we first need to look at the acceleration. Remember Newton's second law of motion? Force equals mass times acceleration ($F=ma$).

So, the acceleration of our particle at any instant is simply the force at that instant divided by its mass. In terms of kinematics, acceleration is the rate of change of velocity, or the derivative of velocity with respect to time, $\frac{dv}{dt}$. This relationship is our master key to unlock the velocity function.

Let's substitute our specific force into this relationship. We get:

Notice how the acceleration depends on time. Because of that negative exponent, as time increases, the acceleration decreases. Geometrically, the acceleration is the slope of the velocity-time graph. So, we are looking for a graph where the slope starts off steep and gradually flattens out.

Now, to find the actual velocity function, we need to solve this differential equation. We can do this by separating the variables. Let's multiply both sides by $dt$ to get all the time terms on the right side. Then, we integrate both sides.

The particle starts from rest, so at time $t=0$, velocity $v=0$. These are our lower limits. Our upper limits will be some arbitrary time $t$ and the corresponding velocity $v$.

Let's execute the integration. On the left side, the integral of $dv$ is simply $v$. On the right side, $\frac{F_0}{m}$ is a constant, so it stays outside. The integral of $e^{-bt}$ is $\frac{e^{-bt}}{-b}$.

Applying the upper limit $t$, we get $e^{-bt}$. Subtracting the lower limit, we plug in zero to get $e^0$, which is exactly $1$.

We can absorb that negative sign from the denominator into the bracket, flipping the terms around. This gives us our beautiful final expression:

Let's analyze this final equation to pick the right graph. At $t=0$, $v=0$.

As time approaches infinity ($t\to \infty$), the term $e^{-bt}$ approaches zero. This means the velocity doesn't increase forever; it approaches a maximum constant value:

This represents a horizontal asymptote. The velocity increases, but its rate of increase slows down over time. This perfectly describes a concave-down curve that levels off. Looking at our options, graph (c) is the exact match.