Understanding Velocity and Acceleration of a Moving Particle

In this article, we will explore the motion of a particle moving along a straight line, focusing on its velocity and acceleration at specific time points. Given the displacement function s(t) t^3 - 12t^2 45t 3, we will compute the velocity and acceleration, and determine when the velocity reaches its maximum and minimum.

Displacement, Velocity, and Acceleration Functions

The displacement of the particle is described by the function s(t) t^3 - 12t^2 45t 3. To find the velocity, we need to compute the first derivative of the displacement function with respect to time, t.

Velocity Function: v(t) frac{d}{dt} s(t) 3t^2 - 24t 45

The acceleration function, on the other hand, is the first derivative of the velocity function, which is the second derivative of the displacement function.

a(t) frac{d^2}{dt^2} s(t) 6t - 24

Identifying Critical Points

To find the times at which the velocity reaches a maximum or minimum, we need to determine the critical points of the velocity function. These are found by setting the velocity function equal to zero and solving for t.

3t^2 - 24t 45 0

Dividing the entire equation by 3 simplifies the problem:

t^2 - 8t 15 0

Factoring the quadratic equation:

(t - 3)(t - 5) 0

Thus, the critical points are: t 3 and t 5.

Second Derivative Test

To determine if these critical points correspond to a maximum or minimum, we use the second derivative test. We evaluate the acceleration function, which is the second derivative of the displacement function, at these critical points.

a(t) 6t - 24

Evaluating at t 3:

a(3) 6(3) - 24 18 - 24 -6

Evaluating at t 5:

a(5) 6(5) - 24 30 - 24 6

Since the acceleration is negative at t 3, the velocity reaches a maximum at this point. Conversely, since the acceleration is positive at t 5, the velocity reaches a minimum at this point.

Velocity and Acceleration at Specific Time Points

Let's now consider the specific time point of t 2. We compute the velocity at this point:

v(2) 3(2)^2 - 24(2) 45 12 - 48 45 9 text{ m/s}

We also compute the acceleration at t 2:

a(2) 6(2) - 24 12 - 24 -12 text{ m/s}^2

Since the acceleration is negative, the particle is slowing down at t 2 seconds.

Summary

The velocity of the particle reaches a maximum at t 3 seconds and a minimum at t 5 seconds. At t 2 seconds, the velocity is 9 m/s and the particle is experiencing a deceleration of 12 m/s2.