Question

1. A particle is moving with acceleration $a(t) = 6 + 12t - 24t^2$, where $t$ is time in seconds. Find the position $s(t)$ (in meters) if $v(1) = -3$ and $s(0) = 5$:

          1. A particle is moving with acceleration $a(t) = 6 + 12t - 24t^2$, where $t$ is time in seconds. Find the position $s(t)$ (in meters) if $v(1) = -3$ and $s(0) = 5$:

1. A particle is moving with acceleration a(t) = 6 + 12t - 24t^2, where t is time in seconds. Find the position s(t) (in meters) if v(1) = -3 and s(0) = 5:

Added by Frank L.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

09/08/2022

Step 1

Given acceleration, a(t) = 6 + 12t + 24t^2 Integrating with respect to t, we get: v(t) = 6t + 6t^2 - 8t^3 + c Show more…

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Transcript

00:01 We're given this acceleration equation.

00:03 In order to find position, we first need to find velocity, because velocity is the direct integral of acceleration.

00:10 So let's take the integral with respect to t of the acceleration formula.

00:17 This is going to give us velocity on the left, and we have 60 plus.

00:23 This is going to be 12 t squared over 2.

00:27 So 12 over 2 is 6.

00:30 So 6 t squared minus 24 t cubed over 3.

00:36 24 over 3 is 8 t cubed.

00:40 And let's not forget the constant of integration.

00:44 Now we are given the condition that the velocity when the time is 1 is equal to 3.

00:52 So we're going to use this condition to solve for c.

00:55 So it's actually negative 3.

00:58 So negative 3 is equal to 6.

01:01 Let's replace all the t's for 1.

01:04 6 times 1 plus 6 times 1 squared minus 8 times 1 cubed.

01:12 And of course we still have the plus c.

01:15 So minus 3 is equal 6 times 1 times 6 times 1.

01:19 Let's write the numbers first.

01:20 So 6 plus 6 minus 8 plus c.

01:26 So minus 3 is equal to 6.

01:28 Plus 6 is 12.

01:31 Minus 8 is positive 4.

01:36 So now we got plus c...

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