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ad-work [718]
3 years ago
5

The acceleration of an object (in m/s^2) is given by the function a(t) = 9 sin(t). The initial velocity of the object is v(0) =

-11 m/s. a) Find an equation v(t) for the object velocity v(t) =___________b) Find the object's displacement (in meters) from time 0 to time 3 meters c) Find the total distance traveled by the object from time 0 to time 3 meters
Mathematics
1 answer:
pentagon [3]3 years ago
4 0

a) Acceleration is the derivative of velocity. By the fundamental theorem of calculus,

v(t)=v(0)+\displaystyle\int_0^ta(u)\,\mathrm du

so that

v(t)=\left(-11\frac{\rm m}{\rm s}\right)+\int_0^t9\sin u\,\mathrm du

\boxed{v(t)=-\left(2+9\cos t)\right)\frac{\rm m}{\rm s}}

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 s(0)=0\,\mathrm m. Then its displacement over the time interval [0, 3] is

s(0)+\displaystyle\int_0^3v(t)\,\mathrm dt=-\int_0^3(2+9\cos t)\,\mathrm dt=\boxed{-6-9\sin3}

c) The total distance traveled is the integral of the absolute value of the velocity function:

s(0)+\displaystyle\int_0^3|v(t)|\,\mathrm dt

v(t) for 0\le t and v(t)\ge0 for \cos^{-1}\left(-\frac29\right)\le t\le3, so we split the integral into two as

\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}-v(t)\,\mathrm dt+\int_{\cos^{-1}\left(-\frac29\right)}^3v(t)\,\mathrm dt

=\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

\displaystyle=\boxed{2\sqrt{77}-6+4\cos^{-1}\left(-\frac29\right)-9\sin3}

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