Question

The velocity of a particle moving along the x-axis at any time t ≥ 0 is given by D%20cos%20%28%5Cpi%20t%29-%20t%287-2%5Cpi%20%29" id="TexFormula1" title="v(t) = cos (\pi t)- t(7-2\pi )" alt="v(t) = cos (\pi t)- t(7-2\pi )" align="absmiddle" class="latex-formula">
A. Find the particles maximum acceleration over the interval [0,2]. Use derivatives to support and justify your answer.

B. Find the particles minimum velocity over the interval [0,3] Justify your answer.

I know that acceleration is the derivative of velocity and that:
$v'(t)=a(t)=-\pi sin(\pi t)-7+2\pi \\v''(t)=a'(t)=-\pi ^2cos(\pi t)$
But I am confused as to actually figure out the minimum and maximum values using cos and sin

Answer 1

Answer:

A. a(3/2) = 3π − 7

B. v(3) = 6π − 22

Step-by-step explanation:

As you found:

v(t) = cos(πt) − t (7 − 2π)

v'(t) = a(t) = -π sin(πt) − 7 + 2π

v"(t) = a'(t) = -π² cos(πt)

a) When the acceleration is a maximum, a'(t) = 0.

0 = -π² cos(πt)

0 = cos(πt)

πt = π/2 + 2kπ, 3π/2 + 2kπ

πt = π/2 + kπ

t = 1/2 + k

t = 1/2, 3/2

We need to check if these are minimums or maximums.  To do that, we evaluate the sign of a'(t) within the intervals before and after each value.

a'(0) = -π² cos(0) = -π²

a'(1) = -π² cos(π) = π²

a'(2) = -π² cos(2π) = -π²

At t = 1/2, a'(t) changes signs from - to +.  At t = 3/2, a'(t) changes signs from + to -.  Therefore, t = 1/2 is a local minimum and t = 3/2 is a local maximum.  At t = 3/2, the acceleration is:

a(3/2) = -π sin(3π/2) − 7 + 2π

a(3/2) = 3π − 7

Compare to the endpoints:

a(0) = -π sin(0) − 7 + 2π = -7 + 2π

a(2) = -π sin(2π) − 7 + 2π = -7 + 2π

So a(3/2) = 3π − 7 is the global maximum.

b) Use the same steps as before.  When the velocity is a minimum, a(t) = 0.

0 = -π sin(πt) − 7 + 2π

π sin(πt) = -7 + 2π

sin(πt) = -7/π + 2

πt ≈ 3.372 + 2kπ, 6.053 + 2kπ

t ≈ 1.073 + 2k, 1.927 + 2k

t ≈ 1.073, 1.927

Now we evaluate the sign of a(t) in the intervals before and after each value.

a(0) = -π sin(0) − 7 + 2π = -7 + 2π

a(3/2) = -π sin(3π/2) − 7 + 2π = -7 + 3π

a(2) = -π sin(2π) − 7 + 2π = -7 + 2π

At t = 1.073, a(t) changes signs from - to +.  At t = 1.927, a(t) changes signs from + to -.  Therefore. t = 1.073 is a local minimum and t = 1.927 is a local maximum.  At t = 1.073, the velocity is:

v(1.073) = cos(1.073π) − 1.073 (7 − 2π)

v(1.073) = -1.743

Compare to the endpoints:

v(0) = cos(0) − 0 (7 − 2π) = 1

v(3) = cos(3π) − 3 (7 − 2π) = -22 + 6π

Here, v(3) = -22 + 6π is the global minimum.