A particle leaves the origin at \(\mathrm{t}=0\) with an initial velocity \(\overrightarrow{\mathrm{v}}=3 \mathrm{v}_{0} \hat{i} .\) It experiences a constant acceleration \(\vec{a}=-2 a_{1} \hat{i}-5 a_{2} \hat{j} .\) The time at which the particle reaches its maximum x coordinate is \(\left(a_{1}, a_{2} \& v_{0}\right.\) are positive number) (1) \(\frac{3 V_{0}}{10 a_{2}}\) (2) \(\frac{3 V_{0}}{9 a_{1}-5 a_{2}}\) (3) \(\frac{3 V_{0}}{2 a_{1}}\) (4) \(\frac{3 V_{0}}{2 a_{1}+5 a_{2}}\)

Understanding constant acceleration is key in kinematics problems. This concept refers to a state where the acceleration of an object remains unchanged over time.

In the given exercise, the particle experiences a constant acceleration given by \(-2a_{1} \hat{i} - 5a_{2} \hat{j}\). Here, \(a_{1}\) and \(a_{2}\) are constants, meaning that the acceleration doesn't vary as time passes.

When dealing with constant acceleration, the primary equations utilized involve the object's initial velocity, the rate of acceleration, and the elapsed time.

To find the final velocity or displacement of an object under constant acceleration, we use these basic kinematic equations:

• \[ v = u + at \] - This is the standard velocity equation, where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the constant acceleration, and \( t \) is the time.
• \[ s = ut + \frac{1}{2}at^2 \] - This is the displacement equation, where \( s \) is the displacement.

The velocity equation in kinematics helps in determining the velocity of a moving object at any given time. It links initial velocity, acceleration, and time.

In the problem, we use the velocity equation for the x-direction: \[ v_{x} = u_{x} + a_{x}t \]

We are given:

• Initial velocity in the x-direction, \( u_{x} = 3v_{0} \)
• Acceleration in x-direction, \( a_{x} = -2a_{1} \)

Substituting these values, we get:

\[ v_{x} = 3v_{0} - 2a_{1}t \]
This equation allows us to calculate the velocity of the particle at any time \( t \).

To find the time when the particle reaches its maximum x-coordinate, observe that when the particle stops moving in the x-direction (i.e., velocity in the x-direction becomes zero), it has reached its maximum x-coordinate. Therefore, set \( v_{x} = 0 \) and solve for \( t \).

Solving \[ 0 = 3v_{0} - 2a_{1}t \], we find \[ t = \frac{3v_{0}}{2a_{1}} \]

Maximum displacement occurs when the object reaches the farthest point in a particular direction.

In this problem, it is the maximum x-coordinate the particle reaches before changing direction.

From the velocity equation, \( v_{x} = 3v_{0} - 2a_{1}t \), we have determined that the maximum x-coordinate occurs when \( v_{x} = 0 \), yielding \[ t = \frac{3v_{0}}{2a_{1}} \]

This time value is critical because it tells us when the particle stops moving forward and starts moving in the opposite direction due to the negative acceleration.

To find the maximum x-coordinate displacement, we can use the displacement equation in the x-direction: \[ x = u_{x}t + \frac{1}{2}a_{x}t^2 \]
By substituting \( u_{x} = 3v_{0} \) and \( a_{x} = -2a_{1} \) into the equation and using \( t = \frac{3v_{0}}{2a_{1}} \), we get the maximum displacement.

This approach not only helps determine the time and position of maximum displacement but also reinforces the importance of kinematic equations in solving such problems.