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## anonymous 4 years ago A particle moves along the x-axis so that its velocity at any time t > 0 is given by v(t)=(2π − 5)t −sin(πt) . A. Find the acceleration at any time t.

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1. lalaly

the acceleration is the rate of change of velocity, so acceleration is the derivative of velocity$a=\frac{dv}{dt}=(2\pi -5)-\pi \cos \pi t$

2. anonymous

B. Find the minimum acceleration of the particle over the interval [0, 3]. (12 points) So to answer B, I make the derivative = 0?

3. lalaly

to find the minimum acceleration of the particle, looking at 0≤t≤3 see that when t =0, the acceleration is at a minimum. so substitute t=0 in the equation of a

4. anonymous

Thanks, for C would I substitute 2 for t? C. Find the maximum velocity of the particle over the interval [0, 2].

5. anonymous

?

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