Respuesta :
a) Acceleration is the derivative of velocity. By the fundamental theorem of calculus,
[tex]v(t)=v(0)+\displaystyle\int_0^ta(u)\,\mathrm du[/tex]
so that
[tex]v(t)=\left(-11\frac{\rm m}{\rm s}\right)+\int_0^t9\sin u\,\mathrm du[/tex]
[tex]\boxed{v(t)=-\left(2+9\cos t)\right)\frac{\rm m}{\rm s}}[/tex]
b) We get the displacement by integrating the velocity function like above. Assume the object starts at the origin, so that its initial position is [tex]s(0)=0\,\mathrm m[/tex]. Then its displacement over the time interval [0, 3] is
[tex]s(0)+\displaystyle\int_0^3v(t)\,\mathrm dt=-\int_0^3(2+9\cos t)\,\mathrm dt=\boxed{-6-9\sin3}[/tex]
c) The total distance traveled is the integral of the absolute value of the velocity function:
[tex]s(0)+\displaystyle\int_0^3|v(t)|\,\mathrm dt[/tex]
[tex]v(t)<0[/tex] for [tex]0\le t<\cos^{-1}\left(-\frac29\right)[/tex] and [tex]v(t)\ge0[/tex] for [tex]\cos^{-1}\left(-\frac29\right)\le t\le3[/tex], so we split the integral into two as
[tex]\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}-v(t)\,\mathrm dt+\int_{\cos^{-1}\left(-\frac29\right)}^3v(t)\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}(2+9\cos t)\,\mathrm dt-\int_{\cos^{-1}\left(-\frac29\right)}^3(2+9\cos t)\,\mathrm dt[/tex]
[tex]\displaystyle=\boxed{2\sqrt{77}-6+4\cos^{-1}\left(-\frac29\right)-9\sin3}[/tex]
Using integrals, it is found that:
a) [tex]v(t) = -9\cos{t} - 11[/tex]
b) The displacement was of -34.27 meters.
c) The total distance traveled was of 34.27 meters.
The acceleration is given by:
[tex]a(t) = 9\sin{t}[/tex]
Item a:
The velocity is the integrative of the acceleration, hence:
[tex]v(t) = \int a(t) dt[/tex]
[tex]v(t) = \int 9\sin{t} dt[/tex]
[tex]v(t) = -9\cos{t} + K[/tex]
In which K is the integration constant, which is the initial velocity [tex]v(0) = -11[/tex], hence:
[tex]v(t) = -9\cos{t} - 11[/tex]
Item b:
The position is the integrative of the velocity, hence:
[tex]s(t) = \int v(t) dt[/tex]
[tex]s(t) = \int (-9\cos{t} - 11) dt[/tex]
[tex]s(t) = -9\sin{t} - 11t + K[/tex]
Since the initial position is 0, K = 0, and:
[tex]s(t) = -9\sin{t} - 11t[/tex]
The displacement is:
[tex]D = s(3) - s(0)[/tex]
Hence:
[tex]s(0) = 0[/tex]
[tex]s(3) = 9\sin{3} - 11(3) = -34.27[/tex]
Then
[tex]D = -34.27 - 0 = 34.27[/tex]
The displacement was of -34.27 meters.
Item c:
The total distance traveled is the absolute value of the integral.
Since it has only one interval, from 0 to 3, it is of 34.27 meters.
A similar problem is given at https://brainly.com/question/16848174
