# [SOLVED]2.1.299 acceleration

#### karush

##### Well-known member
$\tiny{299}$
For $t \ge 0$ the position of a particle moving along the x-axis is given by $v(t)=\sin t—\cos t$ What is the acceleration of the particle at the point where the velocity is first equal to 0?
$a. \sqrt{2}$
$b. \, —1$
$c. \, 0$
$d. \, 1$
$e. —\sqrt{2}$

Ok well originally it was given as $x(t)$ but I changed it to $v(t)$
So via W|A $v(t)=0$ at
$t = 1/4 (4 π n + π), n \in Z$

So the first 0 would be $\dfrac{5\pi}{4}$

Hopefully so far #### topsquark

##### Well-known member
MHB Math Helper
$\tiny{299}$
For $t \ge 0$ the position of a particle moving along the x-axis is given by $v(t)=\sin t—\cos t$ What is the acceleration of the particle at the point where the velocity is first equal to 0?
$a. \sqrt{2}$
$b. \, —1$
$c. \, 0$
$d. \, 1$
$e. —\sqrt{2}$

Ok well originally it was given as $x(t)$ but I changed it to $v(t)$
So via W|A $v(t)=0$ at
$t = 1/4 (4 π n + π), n \in Z$

So the first 0 would be $\dfrac{5\pi}{4}$

Hopefully so far Wait! What do you mean by you changed it to v(t)? The particle's position is x(t) = sin(t) - cos(t) means that v = $$\displaystyle \dfrac{dx}{dt}$$. You can't just change that in the middle of the problem!

$$\displaystyle x(t) = cos(t) - sin(t)$$

$$\displaystyle v(t) = -sin(t) - cos(t)$$

So
$$\displaystyle v(t_0) = -sin(t_0) - cos(t_0) = 0 \implies sin(t_0) = -cos(t_0) \implies tan(t_0) = -1$$
which first happens at $$\displaystyle t_0 = \dfrac{5 \pi }{4}$$ as you stated.

(You don't need W|A for this. Put away the toys.)

So what's a(t)?

-Dan

#### skeeter

##### Well-known member
MHB Math Helper
$\tiny{299}$
For $t \ge 0$ the position of a particle moving along the x-axis is given by $v(t)=\sin t—\cos t$ What is the acceleration of the particle at the point where the velocity is first equal to 0?
$a. \sqrt{2}$
$b. \, —1$
$c. \, 0$
$d. \, 1$
$e. —\sqrt{2}$

Ok well originally it was given as $x(t)$ but I changed it to $v(t)$
So via W|A $v(t)=0$ at
$t = 1/4 (4 π n + π), n \in Z$

So the first 0 would be $\dfrac{5\pi}{4}$
why did you do that? ...

... if originally, $x = \sin{t}-\cos{t}$, then $v = \cos{t} + \sin{t} = 0 \implies t = \dfrac{3\pi}{4}$

#### karush

##### Well-known member
<font color="#ff0000">why did you do that? ... </font>

... if originally, $x = \sin{t}-\cos{t}$, then $v = \cos{t} + \sin{t} = 0 \implies t = \dfrac{3\pi}{4}$

Sorry I'm late but my favorite WiFi Hangouts told me to leave.

So $a(t)=\sin{t}-\cos(t)$ .
Then plug in
$a\left( \dfrac{3\pi}{4}\right) =\sin{ \dfrac{3\pi}{4}}-\cos \dfrac{3\pi}{4}=\sqrt{2}$

Last edited:

#### skeeter

##### Well-known member
MHB Math Helper
Sorry I'm late but my favorite WiFi Hangouts told me to leave.

So $a(t)=\sin{t}-\cos(t)$ .
Then plug in
$a\left( \dfrac{3\pi}{4}\right) =\sin{ \dfrac{3\pi}{4}}-\cos \dfrac{3\pi}{4}=\sqrt{2}$
$v(t) = \cos{t}+\sin{t} \implies a(t) = \cos{t}-\sin{t} \implies a\left(\dfrac{3\pi}{4}\right) = -\sqrt{2}$