Animated Solution for Physics - Kinematics: Two particles A and B start from the same point and move in the positive $x$-direction. In a time interval of $2.00\text{ s}$ after they start, their velocities $v$ vary with time $t$ as shown in the following figures. What is the maximum separation between the particles during this time interval?

Finding Maximum Separation Using Velocity-Time Graphs

Imagine two runners, Particle A and Particle B, starting a race from the exact same starting line. However, they don't run at a constant speed. Their speeds fluctuate wildly, as shown by their velocity-time graphs. Particle A runs in a step-like pattern, while Particle B runs in a jagged, sawtooth pattern. Our mission is to find the exact moment they are furthest apart and calculate that maximum separation.

Relative Velocity

To solve this, we need to think about relative velocity. The separation between the two particles, let's call it $S(t)$, is simply the difference in their positions: $S(t)=x_A(t)-x_B(t)$.

How do we maximize this separation? In calculus, to find the maximum of a function, we take its derivative and set it to zero. The rate of change of separation is the difference in their velocities:

For the separation to be at its absolute maximum, this rate of change must be zero. This leads us to a beautiful, intuitive conclusion: The maximum separation occurs exactly when their velocities are equal ($v_A=v_B$).

Think about it physically: As long as Particle A is running faster than Particle B ($v_A>v_B$), it is pulling away, and the gap is widening. The very instant Particle B becomes faster than Particle A ($v_B>v_A$), it starts catching up, and the gap begins to shrink. The turning point is when their speeds match.

Instead of writing tedious piecewise algebraic equations for their motions, we can use a powerful geometric shortcut. We know that the area under a velocity-time graph represents displacement. Therefore, the separation between the particles is simply the area between their velocity curves.

Let's overlay the two graphs on the same set of axes. We are looking for the points where the blue line (Particle A) and the red line (Particle B) intersect.

Looking closely, we see two intersections: 1. At $t=1.00\text{ s}$ 2. At $t=1.50\text{ s}$

Let's break the motion down into intervals and calculate the area between the curves.

Interval 1: From $t=0$ to $t=1.00\text{ s}$ During this entire first second, Particle A is moving faster than Particle B. The area between the curves forms a neat triangle. - The base of this triangle is $1.00\text{ s}$. - The height is the difference in velocity at $t=0$, which is $2.00\text{ m/s}$.

The separation gained in this interval is:

Interval 2: From $t=1.00$ to $t=1.50\text{ s}$ Right at $t=1.00\text{ s}$, Particle B's velocity suddenly drops to zero, while Particle A is moving at $1.00\text{ m/s}$. Particle A is faster again! The gap continues to widen until $t=1.50\text{ s}$, where Particle B's velocity catches up to $1.00\text{ m/s}$.

The area between the curves here forms a smaller triangle. - The base is $0.50\text{ s}$ (from $1.00$ to $1.50$). - The height is $1.00\text{ m/s}$.

The additional separation gained is:

To find the total maximum separation, we simply add the separation gained in both intervals together:

After $t=1.50\text{ s}$, Particle B's velocity exceeds Particle A's, and the area between the curves becomes 'negative' relative to Particle A, meaning the gap is closing. Thus, $1.25\text{ m}$ is our absolute maximum separation. By trusting the geometry of the graphs, we bypassed complex algebra and arrived at the solution elegantly!