acceleration - velocity & speed

**jvignacio** · Nov 12th 2008, 05:26 AM

An object moves along the

x-axis with acceleration given by a(t) = sin 2t, t being the time, measured in seconds. At t = 0, its position is given by s = 5, and its velocity at t = 0 is v = −1/2 (that is, it is moving to the left at a speed of 1/2 unit per second).

(a) Where is it after 5 seconds?
(b) What is the farthest left it gets, and when does it first get there?

(c) What is the farthest right it gets, and when does it first get there?

how would i start off by tacking this question.

do i need to integrate sin2t to get the velocity?

**mr fantastic** · Nov 12th 2008, 12:03 PM

Originally Posted by jvignacio

An object moves along the

x-axis with acceleration given by a(t) = sin 2t, t being the time, measured in seconds. At t = 0, its position is given by s = 5, and its velocity at t = 0 is v = −1/2 (that is, it is moving to the left at a speed of 1/2 unit per second).

(a) Where is it after 5 seconds?
(b) What is the farthest left it gets, and when does it first get there?

(c) What is the farthest right it gets, and when does it first get there?

how would i start off by tacking this question.

do i need to integrate sin2t to get the velocity?

Yes. And then integrate velocity to get position.

**Soroban** · Nov 12th 2008, 12:33 PM

Hello, jvignacio!

An object moves along the x-axis with acceleration: . $a(t) \:=\:\sin 2t,\:t$ measured in seconds.

At $t = 0$ , its position is $s = 5,$ and its velocity is $v = \text{-}\tfrac{1}{2}$

(a) Where is it after 5 seconds?

(b) What is the farthest left it gets, and when does it first get there?

(c) What is the farthest right it gets, and when does it first get there?

We are given the acceleration function.
We must find the velocity function and the position function.

To find $v(t)$ , integrate $a(t)\!\quad v(t) \;=\;\int\sin2t\,dt \;=\;\text{-}\tfrac{1}{2}\cos2t + C$ .[1]

We are told that: . $v(0) \:=\:\text{-}\tfrac{1}{2}$
. . So we have: . $\text{-}\tfrac{1}{2}\cos0 + C \:=\:\text{-}\tfrac{1}{2} \quad\Rightarrow\quad C \:=\:0$

. . The velocity function is: . $\boxed{v(t) \;=\;\text{-}\tfrac{1}{2}\cos2t}$

To find $s(t)$ , integrate $v(t)\!:\quad s(t) \;=\;\text{-}\tfrac{1}{2}\int \cos2t\,dt \;=\;\text{-}\tfrac{1}{4}\sin2t + C$

We are told that: . $s(0) = 5$
. . So we have: . $\text{-}\tfrac{1}{4}\sin0 + C \:=\:5 \quad\Rightarrow\quad C \:=\:5$

. . The position function is: . $\boxed{s(t) \;=\;\text{-}\tfrac{1}{4}\sin2t + 5}$

Now you are ready to answer the questions . . .

**jvignacio** · Nov 12th 2008, 12:41 PM

Originally Posted by Soroban

Hello, jvignacio!

We are given the acceleration function.
We must find the velocity function and the position function.

To find $v(t)$ , integrate $a(t)\!\quad v(t) \;=\;\int\sin2t\,dt \;=\;\text{-}\tfrac{1}{2}\cos2t + C$ .[1]

We are told that: . $v(0) \:=\:\text{-}\tfrac{1}{2}$
. . So we have: . $\text{-}\tfrac{1}{2}\cos0 + C \:=\:\text{-}\tfrac{1}{2} \quad\Rightarrow\quad C \:=\:0$

. . The velocity function is: . $v(t) \;=\;\text{-}\tfrac{1}{2}\cos2t$

To find $s(t)$ , integrate $v(t)\!:\quad s(t) \;=\;\text{-}\tfrac{1}{2}\int \cos2t\,dt \;=\;\text{-}\tfrac{1}{4}\sin2t + C$

We are told that: . $s(0) = 5$
. . So we have: . $\text{-}\tfrac{1}{4}\sin0 + C \:=\:5 \quad\Rightarrow\quad C \:=\:5$

much appreciated.
so to find its position at 5 seconds, where do i substitute 5?

**mr fantastic** · Nov 12th 2008, 01:14 PM

Originally Posted by jvignacio

much appreciated.
so to find its position at 5 seconds, where do i substitute 5?

You substitute t = 5 into the rule for position(!)

**jvignacio** · Nov 12th 2008, 01:30 PM

Originally Posted by mr fantastic

You substitute t = 5 into the rule for position(!)

thank you and how do i find the the furthest left and right and when does it arrive at these points ?

**mr fantastic** · Nov 12th 2008, 01:35 PM

Originally Posted by jvignacio

thank you and how do i find the the furthest left and right and when does it arrive at these points ?

You mean how to find the furthest left from s = 5 and the furthest right from s = 5 .....? My advice is to draw a graph of s versus t and interpret it accordingly.

**jvignacio** · Nov 12th 2008, 01:48 PM

Originally Posted by mr fantastic

You mean how to find the furthest left from s = 5 and the furthest right from s = 5 .....? My advice is to draw a graph of s versus t and interpret it accordingly.

okay thank you for the help.
much appreciated.

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