Question

a particle moves along the x-axis so that its velocity at any time t is equal to or > 0 is given by v(t)= (2pie-5)t-sin(piet)

a.find the acceleration at any 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]

Answer 1

I will read that as
v(t) = (2π - 5)t - sin(πt)
= 2πt - 5t - sin(πt)

a(t) = 2π - 5 - πcos(πt)
for a min of a , a ' (t) = 0

π^2 sin(πt) = 0
sin(πt) = 0
πt = 0 or πt = π or πt = 2π
t = 0 , t = 1 , t = 2
a(0) = 2π - 5 - πcos0 = π - 5
a(1) = 2π - 5 - πcosπ = 2π-5+π = 3π-5
a(2) = 2π - 5 - πcos2π = 2π - 5 - π = π - 5
a(3) = ... = 3π - 5

looks like the min is π-5

c) for max velocity, a(t) = 0
2πt - 5t - sin(πt) = 0
sin (πt) = 2πt - 5t
very messy equation to solve,
I ran it through Wolfram's "magic" webpage and got
t = 0 and t = .670349

so
v(.670349) = 0
v(0) = sin0 = 0
v(2) = (2π-5)(2) - sin 2π
- 2.566.. - 0 = 2.566

check my arithmetic