Calculus word problem help

Question

The velocity at time t seconds of an object moving on a line is v(t) = 2cos(πt),with distance measured in meters.

The position of the object at time zero seconds is 2 meters. Find the acceleration function a(t) and the position function s(t).

Answer 1

1) The acceleration function is found by differentiating the velocity function.

2) The location function is found by integrating the velocity function, and using the initial condition to find the constant of integration. Details follow:

1) Taking derivative of v(t) gives, by the Chain Rule, a(t)=(-2sin(pi*t)*pi= -2pi*cos(pi*t).

2) Taking anti-derivative of v(t) gives s(t)= (2/pi)*sin(pi*t) + C, which can be checked by differentiation.

Fitting the initial value s(0)=2 gives

2=(2/pi)*sin(0) + C; but sin (0)=0. So C=2

The final result is then s(t) = (2/pi)*sin(pi*t) + 2

Answer 2

Acceleration is the derivative of velocity therefore (using the chain rule) a(t) = v'(t) = -2sin(πt)*(π) or:

a(t) = -2πsin(πt)

Position is the integral of velocity so s(t) = ∫v(t)dt = ∫2cos(πt) dt = 2∫cos(πt) dt = 2sin(πt)/π + C. Since s(0) = 2, we can solve for C.

2 = 2sin(π(0))/π + C

2 = 2sin(0)/π + C

2 = 2(0)/π + C

2 = 0 + C

C = 2

so:

s(t) = 2sin(πt)/π + 2