Question

A particle moves along a straight line according to s(t) = sin(πt) + 2t^3.

a) Determine the acceleration at t = 4.
b) Is the particle moving towards or away from the origin at t = 1.6?​

Answer 1

The particle's acceleration at t = 4 is 48 m/s². At t = 1.6, the particle's velocity is negative, indicating that it is moving towards the origin.

Step-by-step explanation:

To find the acceleration (a) of a particle at t = 4, we must differentiate the position function s(t) = sin(πt) + 2t^3 twice with respect to time. The first derivative gives us the velocity v(t) = πcos(πt) + 6t^2, and the second derivative gives us the acceleration a(t) = -π^2sin(πt) + 12t. At t = 4, the acceleration is a(4) = -π^2sin(4π) + 12(4) = 48 m/s² since sin(4π) is zero.

For part b), the particle is moving towards or away from the origin depending on the sign of its velocity at t = 1.6. So we calculate v(1.6) = πcos(1.6π) + 6(1.6)^2. Since cos(1.6π) is negative and the term 6(1.6)^2 is positive, but smaller in magnitude than πcos(1.6π), the overall velocity is negative. This indicates that the particle is moving towards the origin at t = 1.6.