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

16A particle moves along the xaxis so that its velocity y at any given time t for O 16 is given byMt e 1 At time t0 the particle is at the origin During what in class=

Respuesta :

We have an expression for the velocity of the particle:

[tex]v(t)=e^{2\sin t}-1[/tex]

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:

[tex]\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}[/tex]

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:

[tex]f=\frac{16}{2\pi}\approx2.546[/tex]

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

We can list the intervals as:

[tex]\begin{gathered} (\pi,2\pi) \\ (3\pi,4\pi) \\ (5\pi,16) \end{gathered}[/tex]

The third period is cut at t = 16.

We can skecth the velocity as:

We can round the intervals to the nearest thousand as:

[tex]\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}[/tex]

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

Ver imagen QuintaviousM635727
ACCESS MORE
EDU ACCESS