Part a The acceleration is the derivative of

Part (a) The acceleration is the derivative of the velocity. a(t) = p/3 cos((p/3)t) a(4) = p/3 cos(4 p/3) a(4) = p/3 [-1/2] = -p/6

Part (b) TRUE or FALSE: For 3<t<4. 5, the velocity of the object is decreasing. We know this because a(t) is negative. a(t) = (p/3) cos ((p/3)t) Always positive v(t) 1 Quad III--Cosine is always negative-1 there. 1 2 3 4

Part (b) TRUE or FALSE: For 3<t<4. 5, the speed of the object is increasing. Speed is the absolute value of velocity. Even though the velocity is decreasing (in this case, getting more negative), the v(t) speed is increasing. This is equivalent to putting your 1 car in reverse and hitting the accelerator. Your velocity would grow more 1 2 but 3 the 4 car would and more negative, continue to-1 move faster (speed). Once again, look at the graph from 3 to 4. 5:

Part (c) TD = 3 0 sin((p/3)t) dt - 4 0 sin((p/3)t) dt 3 0 TD = 1. 910 - -. 477 = 2. 387 To calculate total distance, we need v(t) to graph the function to find out where it 1 is positive & negative. 1 -1 2 3 4

Part (d) To find the position of an object, integrate the velocity equation. x(t) = sin((p/3)t) dt x(t) = 3/p sin u du u = ((p/3)t du = p/3 dt

Part (d) x(t) = -3/p cos (p/3)t + C x(t) = -3/p cos u + C x(t) = 3/p sin u du

Part (d) x(t) = -3/p cos (p/3)t + C x(0) = -3/p cos 0 + C 2 = -3/p + C 2 + 3/p = C

Part (d) x(4) = -3/p cos (4 p/3) + 2 + 3/p x(4) = (-3/p)(-1/2) + 2 + 3/p x(4) = (9/2 p) + 2 2 + 3/p = C x(4) = 3. 432