Askiitians Tutor Team
Last Activity: 8 Days ago
To solve this problem, we need to analyze the motion of the two particles under the influence of gravity while they are moving horizontally. Since they are initially moving in opposite directions, we can track their positions and velocities over time until their velocity vectors become perpendicular to each other.
Understanding the Motion of the Particles
Both particles are subject to uniform acceleration due to gravity, which means their vertical motion can be described by the equations of motion under constant acceleration. The horizontal motion, however, is uniform since there is no horizontal acceleration acting on them. Let's break down the problem step by step.
Initial Conditions
- Particle 1: Initial velocity \( v_1 = 3 \, \text{m/s} \) (to the right)
- Particle 2: Initial velocity \( v_2 = 4 \, \text{m/s} \) (to the left)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)
Velocity Components Over Time
As time progresses, the vertical velocity of each particle will increase due to gravity. The vertical velocity \( v_y \) at any time \( t \) can be calculated using the formula:
Vertical velocity: \( v_y = g \cdot t \)
For the horizontal component, since there is no acceleration, the velocities remain constant:
Horizontal velocity: Particle 1: \( v_{1x} = 3 \, \text{m/s} \), Particle 2: \( v_{2x} = -4 \, \text{m/s} \)
Condition for Perpendicular Velocities
For the velocity vectors to be mutually perpendicular, the dot product of their velocity vectors must equal zero. The velocity vector for each particle can be expressed as:
- Particle 1: \( \vec{v_1} = (3, 10t) \)
- Particle 2: \( \vec{v_2} = (-4, 10t) \)
The dot product is given by:
Dot product: \( \vec{v_1} \cdot \vec{v_2} = 3 \cdot (-4) + (10t) \cdot (10t) = -12 + 100t^2 \)
Setting the dot product to zero for perpendicularity:
Equation: \( -12 + 100t^2 = 0 \)
Solving for Time
Rearranging the equation gives:
Equation: \( 100t^2 = 12 \)
Thus, we find:
Time: \( t^2 = \frac{12}{100} = 0.12 \) leading to \( t = \sqrt{0.12} \approx 0.346 \, \text{s} \)
Calculating the Distance Between the Particles
Now that we have the time when their velocities are perpendicular, we can calculate the distance each particle has traveled horizontally:
- Distance traveled by Particle 1: \( d_1 = v_{1x} \cdot t = 3 \cdot 0.346 \approx 1.038 \, \text{m} \)
- Distance traveled by Particle 2: \( d_2 = v_{2x} \cdot t = -4 \cdot 0.346 \approx -1.384 \, \text{m} \)
The total distance between the two particles is the sum of the absolute values of their horizontal displacements:
Total distance: \( D = |d_1| + |d_2| = 1.038 + 1.384 \approx 2.422 \, \text{m} \)
Final Result
At the moment when the velocity vectors of the two particles become mutually perpendicular, the distance between them is approximately 2.422 meters.
