# Thread: acceleration - velocity & speed

1. ## acceleration - velocity & speed

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?

2. 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.

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

4. 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?

5. 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(!)

6. 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 ?

7. 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.

8. 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.