Answer:
D. ∫₀⁷'⁶ 1/π (sin(πt) + ½) dt + | ∫₇,₆¹¹'⁶ 1/π (sin(πt) + ½) dt |
Step-by-step explanation:
Velocity is the integral of acceleration with respect to time.
v = ∫ a(t) dt
v = ∫ cos(πt) dt
v = 1/π sin(πt) + C
At t = 0, v = 1/(2π).
1/(2π) = 1/π sin(π×0) + C
1/(2π) = C
v(t) = 1/π sin(πt) + 1/(2π)
Distance is the integral of the absolute value of velocity with respect to time.
d = ∫ |v(t)| dt
d = ∫ |1/π sin(πt) + 1/(2π)| dt
We need to find where v(t) is positive/negative. To do that, we find where v(t) is 0.
0 = 1/π sin(πt) + 1/(2π)
-1/π sin(πt) = 1/(2π)
sin(πt) = -1/2
πt = 7π/6, 11π/6
t = 7/6, 11/6
Now evaluate the sign of v(t) in each interval.
Between t=0 and t=7/6, v(t) > 0.
Between t=7/6 and t=11/6, v(t) < 0.
So we can write the distance integral as:
d = | ∫₀⁷'⁶ (1/π sin(πt) + 1/(2π)) dt | + | ∫₇,₆¹¹'⁶ (1/π sin(πt) + 1/(2π)) dt |
Since the first integral is always positive:
d = ∫₀⁷'⁶ (1/π sin(πt) + 1/(2π)) dt + | ∫₇,₆¹¹'⁶ (1/π sin(πt) + 1/(2π)) dt |
Factor out 1/π from both:
d = ∫₀⁷'⁶ 1/π (sin(πt) + ½) dt + | ∫₇,₆¹¹'⁶ 1/π (sin(πt) + ½) dt |
Answer is D.
