Video: AP Calculus AB Exam 1 β€’ Section I β€’ Part B β€’ Question 80

A particle moves along a line so that its acceleration for 𝑑 β‰₯ 0 is given by π‘Ž(𝑑) = (2𝑑 + 1)/(𝑒^(2𝑑 + 1)). If the particle’s velocity at 𝑑 = 1 is 5.264, find the velocity at 𝑑 = 0.

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Video Transcript

A particle moves along a line so that its acceleration for 𝑑 is greater than or equal to zero is given by π‘Ž of 𝑑 equals two 𝑑 plus one over 𝑒 to the two 𝑑 plus one. If the particle’s velocity at 𝑑 equals one is 5.264, find the velocity at 𝑑 equals zero.

We’ve been given the rate of acceleration which is equal to the derivative with respect to 𝑑 of this particle’s velocity. That means we can find 𝑉 of 𝑑 by integrating π‘Ž of 𝑑. 𝑉 of 𝑑 equals the integral of π‘Ž of 𝑑 with respect to 𝑑. And because this is indefinite, we need to add our constant of integration plus 𝑐. At this point, we can note that we’re not asked to find 𝑉 of 𝑑. We’re only interested in the velocity at the particular point 𝑑 equals zero. We’re also given that the velocity at 𝑑 equals one is 5.264. And again, the velocity at zero is what we’re trying to find.

Combining this information and the general formula we started with, the definite integral from zero to one of π‘Ž of 𝑑 with respect to 𝑑 is equal to 𝑉 of one minus 𝑉 of zero. Notice that at this point, we haven’t included our constant of integration. This is because as we rearrange the equation, we make the integral the subject and we have 𝑉 of 𝑑 minus 𝑐. And then make that the definite integral from zero to one which is equal to 𝑉 of one minus 𝑐 minus 𝑉 of zero minus 𝑐. And the two 𝑐s cancel each other out, leaving us with the definite integral from zero to one of π‘Ž of 𝑑 with respect to 𝑑 is equal to 𝑉 of one minus 𝑉 of zero.

Back to the task at hand, we’re interested in the velocity at zero. At this point, we’ll go ahead and plug in some information that we know. We know that the 𝑉 of one is 5.264. So we can subtract 5.264 from both sides. We now need to substitute what π‘Ž of 𝑑 equals: two 𝑑 plus one over 𝑒 to the two 𝑑 plus one. To integrate this function would be very difficult. However, because this is a definite integral, we can use our calculator to calculate this integral from zero to one.

So we plug this into the calculator. And it gives us 0.2683053. Bring the rest of the equation down. Subtract 5.264 from what we found the definite integral was. Our equation now says negative 4.9956947 equals the negative velocity of zero, which we can multiply through by negative one to get the positive velocity at zero equals 4.9956947. And since our first velocity was given to three decimal places to the thousandths place, we’ll also round the velocity of zero to the three decimal places. 𝑉 of zero equals 4.996.

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