A Particle Is Moving Along The { X$}$-axis With Velocity { V(t) = E^t \cos \left(e^t - 1\right)$}$. At Time { T = 0$}$, The Particle Is At { X = 2$}$.a. Show That For Values Of { T$}$ Where [$t
15. A Particle Moving Along the x-axis with a Complex Velocity Function
In this problem, we are given a particle moving along the x-axis with a velocity function that is a combination of exponential and trigonometric functions. The velocity function is given by {v(t) = e^t \cos \left(e^t - 1\right)$}$, and we are asked to show that for values of {t$}$ where {t > 1$}$, the particle is moving in the positive direction.
The velocity function is given by {v(t) = e^t \cos \left(e^t - 1\right)$}$. This function is a combination of an exponential function and a trigonometric function. The exponential function {e^t$}$ is always positive, and the cosine function {\cos \left(e^t - 1\right)$}$ oscillates between -1 and 1.
To determine the direction of the particle, we need to find the sign of the velocity function. The velocity function is positive when {\cos \left(e^t - 1\right) > 0$}$, and negative when {\cos \left(e^t - 1\right) < 0$}$.
To solve for the sign of the velocity function, we need to find the values of {t$}$ for which {\cos \left(e^t - 1\right) > 0$}$ and {\cos \left(e^t - 1\right) < 0$}$.
Case 1: {\cos \left(e^t - 1\right) > 0$}$
When {\cos \left(e^t - 1\right) > 0$}$, the velocity function is positive. This occurs when {e^t - 1 > 0$}$ and {e^t - 1 < \pi$}$.
Solving for {t$}$
To solve for {t$}$, we need to isolate {t$}$ in the inequality {e^t - 1 > 0$}$. This gives us {e^t > 1$}$, which is equivalent to {t > 0$}$.
Solving for {t$}$ (continued)
To solve for {t$}$, we also need to isolate {t$}$ in the inequality {e^t - 1 < \pi$}$. This gives us {e^t < 1 + \pi$}$, which is equivalent to {t < \ln (1 + \pi)$}$.
Combining the Inequalities
Combining the inequalities {t > 0$}$ and {t < \ln (1 + \pi)$}$, we get ${0 < t < \ln (1 + \pi)\$}.
Case 2: {\cos \left(e^t - 1\right) < 0$}$
When {\cos \left(e^t - 1\right) < 0$}$, the velocity function is negative. This occurs when {e^t - 1 < 0$}$ or {e^t - 1 > \pi$}$.
Solving for {t$}$
To solve for {t$}$, we need to isolate {t$}$ in the inequality {e^t - 1 < 0$}$. This gives us {e^t < 1$}$, which is equivalent to {t < 0$}$.
Solving for {t$}$ (continued)
To solve for {t$}$, we also need to isolate {t$}$ in the inequality {e^t - 1 > \pi$}$. This gives us {e^t > 1 + \pi$}$, which is equivalent to {t > \ln (1 + \pi)$}$.
Combining the Inequalities
Combining the inequalities {t < 0$}$ and {t > \ln (1 + \pi)$}$, we get {\ln (1 + \pi) < t < 0$}$.
In conclusion, the velocity function {v(t) = e^t \cos \left(e^t - 1\right)$}$ is positive when ${0 < t < \ln (1 + \pi)\$} and negative when {\ln (1 + \pi) < t < 0$}$.
The position function is given by {x(t) = \int v(t) dt$}$. To find the position function, we need to integrate the velocity function.
To integrate the velocity function, we can use the substitution method. Let {u = e^t - 1$}$. Then {du = e^t dt$}$, and {dt = \frac{1}{e^t} du$}$.
Substituting into the Integral
Substituting into the integral, we get {x(t) = \int e^t \cos \left(e^t - 1\right) \frac{1}{e^t} du$}$.
Simplifying the Integral
Simplifying the integral, we get {x(t) = \int \cos u du$}$.
Evaluating the Integral
Evaluating the integral, we get {x(t) = \sin u + C$}$.
Substituting Back
Substituting back, we get {x(t) = \sin \left(e^t - 1\right) + C$}$.
The initial condition is given by {x(0) = 2$}$. To find the value of {C$}$, we need to substitute {t = 0$}$ into the position function.
Substituting the Initial Condition
Substituting the initial condition, we get ${2 = \sin \left(e^0 - 1\right) + C\$}.
Simplifying the Initial Condition
Simplifying the initial condition, we get ${2 = \sin (-1) + C\$}.
Evaluating the Initial Condition
Evaluating the initial condition, we get ${2 = -\sin 1 + C\$}.
Solving for {C$}$
Solving for {C$}$, we get {C = 2 + \sin 1$}$.
The position function is given by {x(t) = \sin \left(e^t - 1\right) + 2 + \sin 1$}$.
In conclusion, the position function {x(t) = \sin \left(e^t - 1\right) + 2 + \sin 1$}$ satisfies the initial condition {x(0) = 2$}$.
The final answer is {x(t) = \sin \left(e^t - 1\right) + 2 + \sin 1$}$.
Q&A: A Particle Moving Along the x-axis with a Complex Velocity Function
In our previous article, we discussed a particle moving along the x-axis with a velocity function that is a combination of exponential and trigonometric functions. The velocity function is given by {v(t) = e^t \cos \left(e^t - 1\right)$}$, and we showed that for values of {t$}$ where {t > 1$}$, the particle is moving in the positive direction.
Q: What is the velocity function of the particle?
A: The velocity function of the particle is given by {v(t) = e^t \cos \left(e^t - 1\right)$}$.
Q: What is the position function of the particle?
A: The position function of the particle is given by {x(t) = \sin \left(e^t - 1\right) + 2 + \sin 1$}$.
Q: What is the initial condition of the particle?
A: The initial condition of the particle is given by {x(0) = 2$}$.
Q: What is the direction of the particle for values of {t$}$ where {t > 1$}$?
A: The particle is moving in the positive direction for values of {t$}$ where {t > 1$}$.
Q: What is the sign of the velocity function for values of {t$}$ where {t > 1$}$?
A: The velocity function is positive for values of {t$}$ where {t > 1$}$.
Q: What is the range of values of {t$}$ for which the velocity function is positive?
A: The velocity function is positive for values of {t$}$ where ${0 < t < \ln (1 + \pi)\$}.
Q: What is the range of values of {t$}$ for which the velocity function is negative?
A: The velocity function is negative for values of {t$}$ where {\ln (1 + \pi) < t < 0$}$.
Q: What is the final answer for the position function of the particle?
A: The final answer for the position function of the particle is {x(t) = \sin \left(e^t - 1\right) + 2 + \sin 1$}$.
In this article, we answered some common questions related to a particle moving along the x-axis with a complex velocity function. We discussed the velocity function, position function, initial condition, direction of the particle, sign of the velocity function, range of values of {t$}$ for which the velocity function is positive or negative, and the final answer for the position function of the particle.
Q: What is the velocity function of the particle?
A: The velocity function of the particle is given by {v(t) = e^t \cos \left(e^t - 1\right)$}$.
Q: What is the position function of the particle?
A: The position function of the particle is given by {x(t) = \sin \left(e^t - 1\right) + 2 + \sin 1$}$.
Q: What is the initial condition of the particle?
A: The initial condition of the particle is given by {x(0) = 2$}$.
Q: What is the direction of the particle for values of {t$}$ where {t > 1$}$?
A: The particle is moving in the positive direction for values of {t$}$ where {t > 1$}$.
Q: What is the sign of the velocity function for values of {t$}$ where {t > 1$}$?
A: The velocity function is positive for values of {t$}$ where {t > 1$}$.
Q: What is the range of values of {t$}$ for which the velocity function is positive?
A: The velocity function is positive for values of {t$}$ where ${0 < t < \ln (1 + \pi)\$}.
Q: What is the range of values of {t$}$ for which the velocity function is negative?
A: The velocity function is negative for values of {t$}$ where {\ln (1 + \pi) < t < 0$}$.
Q: What is the final answer for the position function of the particle?
A: The final answer for the position function of the particle is {x(t) = \sin \left(e^t - 1\right) + 2 + \sin 1$}$.
In this article, we answered some common questions related to a particle moving along the x-axis with a complex velocity function. We discussed the velocity function, position function, initial condition, direction of the particle, sign of the velocity function, range of values of {t$}$ for which the velocity function is positive or negative, and the final answer for the position function of the particle.