Question

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 constant positive number) (1) \(\frac{3 \mathrm{~V}_{0}}{10 \mathrm{a}_{2}}\) (2) \(\frac{3 V_{0}}{9 a_{1}-5 a_{2}}\) (3) \(\frac{3 V_{0}}{2 a_{1}}\) (4) \(\frac{3 V_{0}}{2 a_{1}+5 a_{2}}\)

Short answer

(3) \( \frac{3 V_{0}}{2 a_{1}} \)

Step-by-step solution

Initial Information

The particle starts from the origin with an initial velocity of \(\textbf{v} = 3v_0 \, \hat{i}\). The constant acceleration is given by \(\textbf{a} = -2a_1 \hat{i} - 5a_2 \hat{j}\). We need to find the time at which the particle reaches its maximum x-coordinate.

Expressing Velocity in x-direction

Velocity in the x-direction as a function of time is expressed as \(v_x(t) = v_{0_x} + a_{x} t\). Here, \(v_{0_x} = 3v_0\) and \(a_x = -2a_1\). So, \(v_x(t) = 3v_0 - 2a_1 t\).

Finding Maximum x-coordinate

The particle reaches its maximum x-coordinate when its velocity in the x-direction becomes zero, i.e., \(v_x(t) = 0\). Setting the equation to zero, we get \(3v_0 - 2a_1 t = 0\).

Solving for Time

Solving for \(t\), we get: \[ 2a_1 t = 3v_0 \] \[ t = \frac{3v_0}{2a_1} \]

Verification with Options

Matching the derived expression with the options given, the correct option is (3) \( \frac{3 V_{0}}{2 a_{1}} \).

Key concepts

Kinematics

Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. In kinematics, we are primarily concerned with quantities such as position, velocity, and acceleration.
For this problem, we focus on the motion of a particle that starts at an origin with a given initial velocity and experiences a constant acceleration. To determine when the particle reaches its maximum x-coordinate, we use kinematic equations to analyze its velocity and displacement over time.
Understanding kinematics helps us break down complex motion problems into manageable parts. We often deal with one-dimensional motion separately for each axis when forces are not aligned along a single line.

Velocity

Velocity is a vector quantity that indicates both the speed of an object and the direction of its motion. Here, the initial velocity is given in the x-direction as \(v_{x0} = 3v_0 \hat{i}\). As the particle moves, its velocity in the x-direction changes due to constant acceleration.
The velocity of the particle in the x-direction over time can be expressed through the equation: \(v_x(t) = v_{x0} + a_{x}t \).
Given:

• Initial velocity in x: \( v_{x0} = 3v_0 \)
• Acceleration in x: \( a_x = -2a_1 \)

Therefore, the velocity in the x-direction at time t becomes: \( v_x(t) = 3v_0 - 2a_1 t \).
This expression allows us to find the time when the particle reaches a certain velocity, including when it stops (velocity equals zero) or when it reaches its maximum displacement.

Acceleration

Acceleration is the rate of change of velocity over time and is also a vector quantity. Here, the particle experiences a constant acceleration of \( \textbf{a} = -2a_1 \hat{i} - 5a_2 \hat{j} \).
This means the acceleration has two components:

• In the x-direction: \( a_x = -2a_1 \)
• In the y-direction: \( a_y = -5a_2 \)

For this problem, we are mainly interested in the x-component because we are finding the maximum x-coordinate. The negative sign in the acceleration indicates that the particle is slowing down in the x-direction due to the opposing acceleration. This slowing down continues until the particle stops moving forward in the x-direction.

Maximum Displacement in Motion

To determine the time at which the particle achieves its maximum displacement in the x-direction, we need to find when its velocity in the x-direction becomes zero. As derived previously, the velocity in the x-direction is \( v_x(t) = 3v_0 - 2a_1 t \).
At maximum displacement, \( v_x(t) = 0 \):
\[ 3v_0 - 2a_1 t = 0 \]
Solving for t:
\[ 2a_1 t = 3v_0 \]
\[ t = \frac{3v_0}{2a_1} \]
This is the time when the particle reaches its maximum x-coordinate. So, at \( t = \frac{3v_0}{2a_1} \), the particle stops moving in the x-direction because its velocity becomes zero. It's crucial to understand that maximum displacement does not mean the particle stops entirely, as it can still move in other directions influenced by other components of its acceleration, in this case, the y-component.