# calculus

posted by on .

A particle is moving along the x- axis so that its velocity at any time t ≥0 is give by v(t) = (2 pi - 5)t - sin(pi t)
A. Find the acceleration at ay time t.
B. Find the minimum acceleration of the particle over the interval [0,3].
C. Find the maximum velocity of the particle over the interval [0,2].

• calculus - ,

a(t) = dv/dt, so
a(t) = (2pi-5) - pi cos(pi t)
as usual, max/min involve derivatives, so max a(t) occurs where da/dt = 0

da/dt = pi^2 sin(pi t)
which is zero at t = 0,1,2,3

how do we know which is min or max? 2nd derivative of a(t), which is
pi^3 cos(pi t)

so, since cos(pi t) > 0 at t=3/2,
min accel is at t=0,2 and is pi-5
max accel is at t=1,3 and is 3pi-5

• calculus - ,

a) a(t) = v'(t) = -2πcos(2πt)

• calculus - ,

ops that's wrong heres right answer

A. a(t) = d/dt (v(t)) = d/dt [(2π-5)t - sin(πt)] = 2π - 5 - πcos(πt)

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B. To find the minimum a(t), we'll derive it and equal to zero:

d/dt (a(t)) = π²sin(πt) = 0 ---> sinπt = 0 ---> πt = 0 + 2k*π, k belongs to the integers

t = 2kπ/π = 2k

Since the v(t) is for any t>0 and the given interval is from 0 to 3, the only possible value for k is 1, therefore t = 2*1 = 2 s.

Minimum a(t) is a(2) = (2π - 5) - πcos(2π) = 2π - 5 - π = π - 5
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C. Max speed: derive v(t) and equal it to zero.

d/dt (v(t)) = a(t) = 0

2π - 5 - πcos(πt) = 0

πcos(πt) = 2π -5

cos(πt) = 2 - (5/π)

πt = acos (2 - 5/π)

t = (1/π)acos(2 - 5/π) ~ 1.63 s

v(0.366) = (2π - 5)*1.63 - sin(π*1.63) ~ 3.01 m/s

DISREGARD MY FIRST POST