Question

hello guys, please help me with my math problem, and it is about derivative.
"If an object has position s(t)= 3sin(2t) at time t, what are its velocity and acceleration when t = (5π)/6.
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I make derivative for both side and get velocity = -(3sqrt3)/2 but my teacher answer is positive, is there something wrong? help me to find its acceleration.
Thank You

Answer 1

ds/dt = velocity. That yields: 6cos(2t). cos(5pi/3) = 0.5, that should come out positive then. Maybe you forgot to multiply it by two?

Answer 2

To get acceleration, take the second derivative, or take the derivative of the velocity. That seems negative, but I am unsure.

Answer 3

find ds/dt

\[\frac{ds}{dy} = 6\cos(2t)\]

next find the 2nd derivative

\[\frac{d^2s}{dt^2} = -12\sin(2t)\]


for velocity substitute 5pi/6 into the 1st derivative

\[v = 6\cos(2\times \frac{5\pi}{6}) = 6\cos(\frac{5\pi}{3})\]

4th quadrant, cost is positive

\[v= 6\times \frac{\sqrt{3}}{2} v = 3\sqrt{3} \]


next for acceleration

note



\[a = -12\sin(2\times \frac{5\pi}{3})\]

4th quadrant sin is negative

\[a = -12 \sin(\frac{5\pi}{3})= -12 \times -\frac{1}{2} = 12\]

Answer 4

my teacher just post the answer but not show steps: v(5π/6) =3 a(5π/6)=6sqr3 O_O

Answer 5

oops got my exact ratios wrong'

cos(5pi/3) = cos(pi/3) = 1/2

\[\sin(5\pi/3) = -\sin(\pi/3) = -\frac{\sqrt{3}}{2}\]

Answer 6

so its 6 x 1/2 = 3

and \[-12 \times \frac{-\sqrt{3}}{2}\]