KINEMATICS O6

Physics Solution

Solution

Analysis of Motion

The maximum separation between two particles moving in the same direction occurs when their relative velocity is zero. That is, when the velocity of the trailing particle equals the velocity of the leading particle ($v_A = v_B$). Before this moment, particle A is moving faster than B, increasing the separation. After this moment, B moves faster than A, reducing the separation.

Step 1: Determine the Time of Maximum Separation

Let’s analyze the velocity-time graphs for both particles:

  • Particle A: Moves with a constant velocity of $2.00 \, \text{m/s}$ for $t \in [0, 1]$ and $1.00 \, \text{m/s}$ for $t \in [1, 2]$.
  • Particle B: Accelerates linearly from $0$ to $2.00 \, \text{m/s}$ during $t \in [0, 1]$. At $t=1$, it resets and accelerates linearly again.

We look for the instant where $v_A = v_B$:

  • In the interval $0 < t < 1$, $v_A = 2$ m/s, while $v_B$ reaches $2$ m/s only at $t=1$.
  • In the interval $1 < t < 2$, $v_A = 1$ m/s. For particle B, the motion is a linear ramp starting from rest at $t=1$ with the same slope as the first interval. The slope (acceleration) is $a_B = \frac{2 - 0}{1 - 0} = 2 \, \text{m/s}^2$. Thus, for $t > 1$, $v_B = 2(t – 1)$.

Equating the velocities for $t > 1$: $$ 1 = 2(t – 1) $$ $$ t – 1 = 0.5 $$ $$ t = 1.5 \, \text{s} $$

t/s v/(m/s) 1.00 2.00 1.00 2.00 Particle A 1.5 t/s v/(m/s) 1.00 2.00 1.00 2.00 Particle B 1.5
Step 2: Calculate Displacement Area ($S_A$ and $S_B$)

The displacement is given by the area under the velocity-time graph. We calculate the area for each particle up to $t = 1.5$ s.

For Particle A:

$$ S_A = \text{Area}(0 \to 1) + \text{Area}(1 \to 1.5) $$ The first part is a rectangle of height 2 and width 1. The second part is a rectangle of height 1 and width 0.5. $$ S_A = (2 \times 1) + (1 \times 0.5) $$ $$ S_A = 2 + 0.5 = 2.5 \, \text{m} $$

For Particle B:

$$ S_B = \text{Area}(0 \to 1) + \text{Area}(1 \to 1.5) $$ The first part is a triangle with base 1 and height 2. $$ \text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 2 = 1 \, \text{m} $$ The second part is a smaller triangle starting at $t=1$. The base is $(1.5 – 1) = 0.5$. The height at $t=1.5$ is $v_B = 1$ m/s. $$ \text{Area}_2 = \frac{1}{2} \times 0.5 \times 1 = 0.25 \, \text{m} $$ $$ S_B = 1 + 0.25 = 1.25 \, \text{m} $$

Step 3: Calculate Maximum Separation

The maximum separation is the difference in displacement at $t=1.5$ s. $$ \Delta x_{\text{max}} = S_A – S_B $$ $$ \Delta x_{\text{max}} = 2.5 – 1.25 $$ $$ \Delta x_{\text{max}} = 1.25 \, \text{m} $$

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