A particle moving along positive X direction passes through origin with a velocity of 5 metre per second when the acceleration is 2 metre per second squared is equals to second the acceleration reverse the direction with a magnitude remaining the same find the maximum displacement of the particle on positive x-axis

To solve this problem, we need to analyze the motion of the particle step by step, taking into account its initial conditions and how the acceleration affects its movement. Let's break it down.

Initial Conditions

The particle starts at the origin (position = 0) with an initial velocity of 5 m/s in the positive x-direction. The initial acceleration is 2 m/s², which means the particle is speeding up in the positive direction.

Phase 1: Positive Acceleration

During the first phase, the particle accelerates positively. We can use the kinematic equation to find the displacement during this phase:

• Displacement (s) = ut + (1/2)at²

Here, u is the initial velocity (5 m/s), a is the acceleration (2 m/s²), and t is the time. However, we need to find the time until the acceleration changes direction.

Finding the Time Until Acceleration Changes

The acceleration reverses direction after a certain time. To find this time, we can use the formula for final velocity:

• v = u + at

At the moment the acceleration reverses, we can assume the final velocity will be zero (the particle stops before reversing direction). Setting v = 0, we have:

• 0 = 5 + (2)t

Solving for t gives:

• t = -5/2 = -2.5 seconds (not physically meaningful in this context)

Instead, we can consider the time until the particle reaches its maximum velocity before the acceleration reverses. The particle will continue to accelerate until it reaches a point where the acceleration is no longer effective in the positive direction.

Phase 2: Negative Acceleration

Once the acceleration reverses, it becomes -2 m/s². The particle will now decelerate until it comes to a stop. The new initial conditions for this phase are:

• Initial velocity (u) = 5 m/s
• Acceleration (a) = -2 m/s²

Finding the Time to Stop

Using the same formula for final velocity:

• 0 = 5 - 2t

Solving for t gives:

• t = 5/2 = 2.5 seconds

Calculating Displacement During Deceleration

Now we can find the displacement during this deceleration phase:

• s = ut + (1/2)at²

Substituting the values:

• s = (5)(2.5) + (1/2)(-2)(2.5)²
• s = 12.5 - 6.25 = 6.25 meters

Maximum Displacement on the Positive X-Axis

Now, we need to add the displacement from both phases to find the total maximum displacement:

• Displacement during positive acceleration: 6.25 meters
• Displacement during negative acceleration: 6.25 meters

Thus, the total maximum displacement of the particle on the positive x-axis is:

• Total Displacement = 6.25 + 6.25 = 12.5 meters

In summary, the maximum displacement of the particle on the positive x-axis is 12.5 meters. This analysis illustrates how the particle's motion is influenced by changes in acceleration and how to apply kinematic equations effectively in different phases of motion.