Question

A particle moves on a horizontal trajectory to the right from the origin. The

velocity equation of the particle is
v() = -4² + 8t + 12 . At t=0, the particle is at position s=4. Determine s(t) and a(t).
Investigate when the particle accelerates/decelerates within the time interval of [0 , 6 ].

Answer 1

To find s(t) and a(t), we integrate and differentiate the given velocity function respectively. s(t) is found to be (-4/3)t³ + 4t² + 12t + 4, while a(t) is -8t + 8. The particle accelerates when a(t) > 0 and decelerates when a(t) < 0.

Step-by-step explanation:

The student has provided the velocity equation of a particle moving along a horizontal trajectory as v(t) = -4t² + 8t + 12. To determine the position function s(t) and the acceleration function a(t), we must integrate and differentiate the velocity function, respectively.

Let's start by finding the acceleration by differentiating the velocity function:

a(t) = d/dt (-4t² + 8t + 12) = -8t + 8.

Next, to find the position function, we integrate the velocity function:

s(t) = ∫ v(t) dt = ∫ (-4t² + 8t + 12) dt.

Don't forget to add the constant of integration, which we can determine using the initial condition s(0) = 4:

s(t) = (-4/3)t³ + 4t² + 12t + C, where C = 4.

The full position equation thus becomes:

s(t) = (-4/3)t³ + 4t² + 12t + 4.

To investigate acceleration and deceleration, we consider the sign of a(t). When a(t) is positive, the particle accelerates; when it is negative, the particle decelerates. Here, the particle accelerates for t < 1s and decelerates for t > 1s within the given interval [0, 6].