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The acceleration function (in $ m/s^2 $) and the initial velocity are given for a particle moving along a line. Find (a) the velocity at time $ t $ and (b) the distance traveled during the given time interval.

$ a(t) = t + 4 $, $ v(0) = 5 $, $ 0 \le t \le 10 $

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a) $\frac{t^{2}}{2}+4 t+5$b)$\frac{1250}{3}(\mathrm{~m})$

03:28

Frank Lin

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

Missouri State University

Oregon State University

University of Nottingham

Boston College

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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The acceleration function …

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The velocity function (in …

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A particle moves with acce…

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So we're given the function of acceleration to be T. Is equal to you. D plus four. Were also given that the velocity is equal to five at time zero. As well as the interval of time being 0-10 seconds. So for part we want to find the velocity at time T in order to find the velocity with are given information. The first thing we want to do is find the integral of our acceleration function. The integral of acceleration will give us velocity. So we have the integral uh T plus four DT. This will end up giving us t squared over two plus 14. Quite see. Can't forget your plus C. Mhm. So next now that we know that the velocity is equal to five at time zero. We're going to plug in zero for everywhere where we see A. T. So we have 0/2 Plus four times 0 plus C. At time zero. So this will end up equaling see and since we know that uh time zero philosophy, 0 to 5 C will end up equaling five. So now from here we can get our velocity with respect to time function. The equal t squared over two plus 14 plus five. And this will be our function for velocity. So next up for part B we want to find the distance traveled during the time integral time interval. So in order to do that first we're going to want to take the integral uh Our time interval 0-10 with respect to our velocity function. The integral of velocity is distance. So B. F T D T. So we're simply going to plug in our velocity function here into our integral. So we have 0-10 of t squared over two plus 40 plus five. Yeah. Okay, so first step will be to evaluate the integral and once we evaluate our integral you should get Teak You'd ever six plus to t squared plus five T. And we want to evaluate this integral once again on the time interval from 0 to 10. So since we have 0-10, we're going to plug in 10 everywhere where we see our T. And we can ignore the second half. We're plugging in zero because when we plug in zero, the whole function is going to equal zero. So you end up with 10 cubed over six plus two times 10 squared Plus five times 10. And when you evaluate this out, You should have 1000 over six plus 200 plus 50. When you evaluate this out, finally, your final answer will end up being about 416 0.67. And this will be the distance that was traveled.

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