# Acceleration @ $\dfrac{5\pi}{4}: \sqrt{2}$

• MHB
• karush
In summary, the position of a particle moving along the x-axis is given by $v(t)=\sin t—\cos t$. The acceleration of the particle at the point where the velocity is first equal to 0 is $\sqrt{2}$ when $v(t)=0$ at $t=\dfrac{5\pi}{4}$ or $a(t)=-\sqrt{2}$ when $v(t)=0$ at $t=\dfrac{3\pi}{4}$.
karush
Gold Member
MHB
$\tiny{299}$
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 🕶

karush said:
$\tiny{299}$
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

karush said:
$\tiny{299}$
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 $v = \cos{t} + \sin{t} = 0 \implies t = \dfrac{3\pi}{4}$

skeeter said:
<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:
karush said:
Sorry I'm late but my favorite WiFi Hangouts told me to leave.So $a(t)=\sin{t}-\cos(t)$ .
Then plug in
$a\left( \dfrac{3\pi}{4}\right) =\sin{ \dfrac{3\pi}{4}}-\cos \dfrac{3\pi}{4}=\sqrt{2}$

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

## 1. What is the formula for acceleration?

The formula for acceleration is a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval.

## 2. How is acceleration measured?

Acceleration is typically measured in meters per second squared (m/s²) or feet per second squared (ft/s²).

## 3. What does "Acceleration @ $\dfrac{5\pi}{4}: \sqrt{2}$" mean?

This notation represents the acceleration at a specific angle and magnitude. In this case, it means that the acceleration has a magnitude of $\sqrt{2}$ and is directed at an angle of $\dfrac{5\pi}{4}$ radians.

## 4. How does acceleration affect an object's motion?

Acceleration is a measure of how quickly an object's velocity changes. If an object is accelerating, it will either speed up, slow down, or change direction.

## 5. Can acceleration be negative?

Yes, acceleration can be negative. This means that the object is decelerating, or slowing down. It can also indicate that the object is moving in the opposite direction of its initial velocity.

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