This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. The acceleration function (in m/s2) and the initial velocity v(0) are given for a particle moving along a line. a(t) = 2t + 4, v(0) = −32, 0 ≤ t ≤ 6 Exercise (a) Find the velocity at time t. Step 1 The velocity function is the antiderivative of the acceleration. v(t) = (2t + 4) dt = $$ Correct: Your answer is correct. webMathematica generated answer key + C Step 2 We must determine the value of C. We know that v(0) = −32. Substituting 0 into our antiderivative gives −32 = v(0) = 0 Correct: Your answer is correct. seenKey 0 + C. Therefore, C = -32 Correct: Your answer is correct. seenKey -32 . Step 3 Therefore, what is the velocity function at time t? v(t) = $$ Your answer is correct. t^2+4t-32 m/s(b) Find the distance traveled during the given time interval. Step 1 The velocity function is v(t) = t2 + 4t − 32, and so the distance traveled in the time interval 0 ≤ t ≤ 6 is given by 6 |t2 + 4t − 32| dt 0 . Remembering that |z| = Incorrect: Your answer is incorrect. z ≥ 0 Incorrect: Your answer is incorrect. z < 0 , we must determine where v(t) = t2 + 4t − 32 is positive or negative. v(t) can be factored as t2 + 4t − 32 = t + 8