Asked by Sarah
The velocity of a particle moving along the x-axis at any time t >= 0 is given by v(t) = cos(pi t) - t(7 - 2pi).
a. Find the acceleration at any time t.
b Find the particle's maximum acceleration over the interval [0, 2]. Use derivatives to support and justify your answer.
c. Find the particle's minimum velocity over the interval [0, 3]. Justify your answer.
Please show all work, all help is greatly appreciated.
a. Find the acceleration at any time t.
b Find the particle's maximum acceleration over the interval [0, 2]. Use derivatives to support and justify your answer.
c. Find the particle's minimum velocity over the interval [0, 3]. Justify your answer.
Please show all work, all help is greatly appreciated.
Answers
Answered by
Damon
v(t) = cos(pi t) - t(7 - 2pi)
Please check for typo in t(7 - 2pi)
As written it is not part of your problem.
a = dv/dt = - pi sin (pi t)
when is a max?
when da/dt = 0
da/dt = -pi^2 cos pi t = 0 at max
cos is 0 when pi t = pi/2 and when pi t = 3 pi/2 and v is positive
but at t = 1/2 a is negative
so at t = 3/2
a max = - pi sin (pi t)= - pi sin (3pi/2) = - pi(-1) = +pi at max
do c the same way. velocity is min when a = zero and v is negative
Please check for typo in t(7 - 2pi)
As written it is not part of your problem.
a = dv/dt = - pi sin (pi t)
when is a max?
when da/dt = 0
da/dt = -pi^2 cos pi t = 0 at max
cos is 0 when pi t = pi/2 and when pi t = 3 pi/2 and v is positive
but at t = 1/2 a is negative
so at t = 3/2
a max = - pi sin (pi t)= - pi sin (3pi/2) = - pi(-1) = +pi at max
do c the same way. velocity is min when a = zero and v is negative
Answered by
Damon
Unless that was not a typo and your velocity keeps diving forever due to that weird
t(7 - 2pi)
t(7 - 2pi)
Answered by
Sarah
That's not a typo, that's just how the question is written.
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