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.
