Analyzing s(t)=t^3-3t: When is It Speeding Up?

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The discussion revolves around determining when the object described by the function s(t)=t^3-3t is speeding up. It is established that the object speeds up when the acceleration, represented by the second derivative s''(t), is positive. The critical points are identified at t=sqrt(2) and t=0, leading to the conclusion that the object is speeding up in the intervals (-sqrt(2), 0) and (sqrt(2), ∞). A mistake is noted regarding the derivative of 3t, which does not affect the overall conclusion about acceleration but does impact the velocity intervals. The final clarification emphasizes the importance of correctly identifying both acceleration and velocity to accurately determine when the object is speeding up.
UrbanXrisis
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s(t)=t^3-3t

which interval(s) of t is this object speeding up?

s'(t)=3x^2-6
0=3x^2-6
x=sqrt(2) and

...+...-...-...+
inf...-sqrt(2)...0...sqrt(2) ...inf

s''(t)=6x
0=6x
x=0

...-...+...
inf...0...inf


inc (-sqrt(2),0) V (sqrt(2),inf)

correct?
 
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UrbanXrisis said:
s(t)=t^3-3t

which interval(s) of t is this object speeding up?

s'(t)=3x^2-6
0=3x^2-6
x=sqrt(2) and
Why did you change your t's to x's. :confused:

The object is speeding up when s''(t) is positive.
6t>0 iff t>0
 
You made one mistake. Derivative of 3t is 3, not 6. That changes the interval where the velocity is positive or negative. Fortunately, that doesn't change your answer as to when acceleration is negative or positive (both 3 and 6 have a derivative of 0).

What's the question actually asking. You asked when the object was speeding up (i.e. - acceleration was positive). Your final answer gave an incorrect answer as to when the object was moving forward (velocity positive).
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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