Question

For t ≥ 0, the velocity of a particle moving along the x-axis is given by v(t)=cos(8e⁻⁰.²ᵗ). The initial position of the particle at t=0 is x=2.5. What is the displacement of the particle from t=0 to t=15?

A. sin(8e⁻⁰.²ᵗ)+C, where C is a constant

B. 1​/8e⁻⁰.²ᵗ cos(8e⁻⁰.²ᵗ)+C, where C is a constant

C. 1/8 ​sin(8e⁻⁰.²ᵗ)+C, where C is a constant

D. 2.5 − 1/8​sin(8e⁻⁰.²ᵗ)

Answer 1

Integrate the given velocity function and apply the initial position to find C. Then, evaluate the resultant function at t=15 to find the displacement of the particle.

Step-by-step explanation:

To find the displacement of the particle from t=0 to t=15, we need to integrate the velocity function, v(t) = cos(8e-0.2t). Given that the initial position is x=2.5 at t=0, the displacement x(t) can be found by integration:

1. Integrate the velocity function to get the position function x(t).
2. Apply the initial condition x(0) = 2.5 to solve for the constant of integration.
3. Calculate the position at t=15 using the position function.

The integral of cos(8e-0.2t) is (1/8)e-0.2t sin(8e-0.2t) plus the constant of integration, C. Substituting the initial condition will help us find C. Therefore, it is important to evaluate the integral correctly and apply the initial conditions.