You could interpret them as system of equation and answer the question easily

from A

-4x + y + x + w =1

write in same form from A

as this system has five variables and four of them are unknowns it means the unique solution is not possible

So, det (A) = 0

8 0

3 years ago

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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) =

pentagon [3]

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

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

$=\displaystyle\int_0^{\cos^{-1}\left(-\frac29\right)}(2+9\cos t)\,\mathrm dt-\int_{\cos^{-1}\left(-\frac29\right)}^3(2+9\cos t)\,\mathrm dt$

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

4 0

3 years ago