Question

A Particle travels on the x-axis so its velocity at time t is given by v(t)=1/2-sin(t), where t is between 0 and 2pi, inclusive.
A. For what values of t is the particle moving to the right?
B. If the particle starts at x=3, what is the final position of the article?
C. What is the total distance traveled by the particle?
D. When t=pi/4, is the speed of the particle increasing or decreasing?

Answer 1

"Moving to the right" corresponds to the particle's velocity being positive, which would indicate an increase in the particle's displacement. This means that part (a) is essentially asking you to find the interval(s) over which \(v(t)\) is positive, i.e. when \(s(t)\) (the displacement) is increasing. This is an application of the first derivative test.

For part (b), you solve the ordinary differential equation,
\[v(t)=s'(t)=\frac{1}{2}-\sin t\]
This is a fairly simple ODE which can be solved by integrating both sides:
\[s(t)=\int s'(t)\,dt=\int\left(\frac{1}{2}-\sin t\right)\,dt+C\]
where you can determine the particular value of \(C\) using the initial value, \(s(0)=3\).

The total distance traveled by the particle is given by the definite integral,
\[\text{distance}=\int_0^{2\pi}|s(t)|\,dt\]
which will require you to split up the domain of integration for negative and positive values of \(s(t)\) such that
\[\text{distance}=\int_{A}s(t)\,dt-\int_Bs(t)\,dt\]
where \(A\) is the subset of the real line over which \(s(t)>0\) and \(B\) is the subset over which \(s(t)<0\).

Part (d) involves taking the derivative of the velocity function and finding the acceleration function. Velocity is increasing when the acceleration is positive and decreasing when negative.