Question

16A particle moves along the x-axis so that its velocity y at any given time t, for O≤† ≤16, is given byM(t) = e-*' -1. At time t=0, the particle is at the origin. During what intervals of time is the particlemoving to the left?Round to the nearest thousandth

Answer 1

We have an expression for the velocity of the particle:

$v(t)=e^{2\sin t}-1$

We have to find when the particle moves to the left.

As the particle moves along the x-axis, this means that the velocity is negative.

We then have to find the interval where v(t) < 0.

We can find this interval as:

$\begin{gathered} v(t)<0 \\ e^{2\sin t}-1<0 \\ e^{2\sin t}<1 \\ \ln(e^{2\sin t})<\ln(1) \\ 2\sin t<0 \\ \sin t<0 \end{gathered}$

The function sin(t) is negative for intervals between π and 2π per cycle.

As t is defined from 0 to 16, we can calculate how many cycles we have:

$f=\frac{16}{2\pi}\approx2.546$

We will have at least 2 intervals or 3 at most where sin(t) < 0.

We can list the intervals as:

$\begin{gathered} (\pi,2\pi) \\ (3\pi,4\pi) \\ (5\pi,16) \end{gathered}$

The third period is cut at t = 16.

We can skecth the velocity as:

We can round the intervals to the nearest thousand as:

$\begin{gathered} (\pi,2\pi)=(3.142,6.286) \\ (3\pi,4\pi)=(9.425,12.566) \\ (5\pi,16)=(15.708,16) \end{gathered}$

Answer: the intervals for t are (3.142, 6.286), (9.425, 12.566) and (15.708, 16).