the position of a particle w.r.t time t along x-axis is given by x=9t^2-t^3

To analyze the motion of a particle along the x-axis described by the equation \( x = 9t^2 - t^3 \), we can derive several important characteristics, such as its velocity and acceleration, which will help us understand how the particle moves over time.

Understanding the Position Function

The position function \( x(t) = 9t^2 - t^3 \) is a polynomial equation where \( t \) represents time. The terms indicate that the particle's position changes with time in a non-linear manner due to the presence of the cubic term \( -t^3 \).

Finding Velocity

Velocity is the rate of change of position with respect to time. To find the velocity function \( v(t) \), we differentiate the position function with respect to \( t \):

• Start with the position function: \( x(t) = 9t^2 - t^3 \)
• Differentiate: \( v(t) = \frac{dx}{dt} = \frac{d}{dt}(9t^2) - \frac{d}{dt}(t^3) \)
• Calculating the derivatives gives us: \( v(t) = 18t - 3t^2 \)

This velocity function tells us how fast the particle is moving at any given time \( t \). The term \( 18t \) indicates that the velocity increases linearly with time, while \( -3t^2 \) introduces a deceleration effect as time progresses.

Determining Acceleration

Acceleration is the rate of change of velocity with respect to time. To find the acceleration function \( a(t) \), we differentiate the velocity function:

• Start with the velocity function: \( v(t) = 18t - 3t^2 \)
• Differentiate: \( a(t) = \frac{dv}{dt} = \frac{d}{dt}(18t) - \frac{d}{dt}(3t^2) \)
• This results in: \( a(t) = 18 - 6t \)

The acceleration function \( a(t) = 18 - 6t \) shows that the acceleration decreases linearly over time. Initially, when \( t = 0 \), the acceleration is positive (18), indicating that the particle is speeding up. However, as \( t \) increases, the acceleration becomes negative, suggesting that the particle will eventually slow down.

Analyzing Motion Characteristics

To gain further insights into the particle's motion, we can find the points where the velocity is zero, which indicates when the particle changes direction:

• Set the velocity function to zero: \( 18t - 3t^2 = 0 \)
• Factoring gives: \( 3t(6 - t) = 0 \)
• This results in \( t = 0 \) and \( t = 6 \) seconds.

At \( t = 0 \), the particle starts from rest, and at \( t = 6 \), it comes to a stop before reversing direction. We can also evaluate the position at these critical points:

• At \( t = 0 \): \( x(0) = 9(0)^2 - (0)^3 = 0 \)
• At \( t = 6 \): \( x(6) = 9(6)^2 - (6)^3 = 324 - 216 = 108 \)

This tells us that the particle starts at the origin and reaches a position of 108 units along the x-axis before reversing direction.

Summary of Motion

In summary, the particle's motion can be characterized as follows:

• Starts at the origin (0,0).
• Reaches a maximum position of 108 units at \( t = 6 \) seconds.
• Changes direction after \( t = 6 \) seconds as the velocity becomes negative.

This analysis provides a comprehensive view of the particle's motion along the x-axis over time, illustrating how its position, velocity, and acceleration interact dynamically. Understanding these concepts is crucial in physics, particularly in kinematics, where we study the motion of objects.