Final answer:
The velocity function v(t) = pt³/3 - qt´/4 and the position function x(t) = pt´/12 - qtµ/20 are found by integrating the acceleration function a(t) = pt² - qt³, with initial conditions of zero velocity and position. Therefore the correct options a) Option a: v(t) = pt^3/3 - qt^4/4; Option b: x(t) = pt^4/12 - qt^5/20.
Step-by-step explanation:
The student is asking about how to find the velocity and position as functions of time for a particle whose acceleration varies with time. Since velocity is the integral of acceleration and position is the integral of velocity, we need to integrate the acceleration function given, a(t) = pt² - qt³. Given that the initial conditions for both velocity and position are zero, we can integrate to find:
Velocity as a Function of Time
∫ a(t) dt = ∫ (pt² - qt³) dt = ∨ pt³/3 - qt´/4 + C
Given that the initial velocity is zero, we find that C = 0. Hence, v(t) = pt³/3 - qt´/4.
Position as a Function of Time
∫ v(t) dt = ∫ (pt³/3 - qt´/4) dt = pt´/12 - qtµ/20 + C
Since the initial position is zero, C = 0. Therefore, x(t) = pt´/12 - qtµ/20.
Based on the above calculations, the correct options for the velocity and position functions of time are given by Option a) v(t) = pt³/3 - qt´/4; and Option b) x(t) = pt´/12 - qtµ/20.