Question

The acceleration of a particle is defined by the relation a = 3e⁻⁰. ²ᵗ, where a and t are expressed in ft/s² and seconds, respectively. Knowing that x = 0 and v = 0 at t = 0, determine the velocity and position of the particle when t = 0.5 s. (Round the final answers to three decimal places.) The velocity and position of the particle when t = 0.5 s is ft/s and ft, respectively.

Answer 1

The velocity and position of a particle given an exponential acceleration function are found by integrating the acceleration with respect to time to obtain the velocity and then integrating the velocity to find the position, both at the specified time of t = 0.5 s.

Step-by-step explanation:

The student is asking for the velocity and position of a particle at time t = 0.5 s, given that the acceleration a of the particle is a function of time t defined by the equation a = 3e−0.2t, and that both the initial velocity v and the initial position x are 0 when t = 0.

To find the velocity at t = 0.5 s, we integrate the acceleration function with respect to time:

1. \( v(t) = \int a(t) \, \mathrm{d}t = \int 3e^{-0.2t} \, \mathrm{d}t \).
2. After integrating, we find \( v(t) = -15e^{-0.2t} + C \), where C is the constant of integration.
3. Since the initial velocity v(0) is 0, we determine C = 15, and hence the velocity function becomes \( v(t) = -15e^{-0.2t} + 15 \).
4. Substituting t = 0.5 s into this equation gives us the velocity at t = 0.5 s.

To find the position at t = 0.5 s, we integrate the velocity function:

1. \( x(t) = \int v(t) \, \mathrm{d}t = \int (-15e^{-0.2t} + 15) \, \mathrm{d}t \).
2. After integrating, we get \( x(t) = -75e^{-0.2t} - 15t + D \), where D is the constant of integration.
3. Since the initial position x(0) is 0, we determine D to be 75, and thus the position function becomes \( x(t) = -75e^{-0.2t} - 15t + 75 \).
4. Substituting t = 0.5 s into this equation gives us the position at t = 0.5 s.

The calculated values need to be rounded to three decimal places to meet the student's request.