Question

The acceleration of an object is given by a(t)=cos(πt). The velocity of the object at time t=0 is 1/(2π). Which expression can be used to find the total distance traveled in the first 11/6 seconds?

Answer 1

Answer:

D

Step-by-step explanation:

v(t) = (1/pi) × sin(pi × t) + c

At t = 0, v = 1/2pi

1/2pi = 0 + c

v(t) = (1/pi)sin(pi×t) + 1/(2pi)

Takes a turn when v = 0

0 = (1/pi)sin(pi×t) - 1/(2pi)

sin(pi × t) = -½

pi × t = 7pi/6

t = 7/6

For t = 0 to t = 7/6,

V(t) > 0

So s(t) > 0

For t = 7/6 to t = 11/6,

Displacement is negative, so put a mod

Answer 2

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.