# Thread: Distance in Respect to Time

1. ## Distance in Respect to Time

A Particle moves along the x axis so that at time t its position is given by x(t)=sin( pi t^2)
(A.)Find the Velocity at time t
(B.)Find the acceleration of time t
(C.)For What Value of t does the particle change direction
(D.)All Values of t that the particle is moving to the left

2. Originally Posted by Dragon
A Particle moves along the x axis so that at time t its position is given by x(t)=sin( pi t^2)
(A.)Find the Velocity at time t
(B.)Find the acceleration of time t
(C.)For What Value of t does the particle change direction
(D.)All Values of t that the particle is moving to the left
a) The velocity is the derivative of position.

b) The acceleration is the derivative of velocity, or the second derivative of position.

c) Find the turning point by setting the velocity to 0 and solving for t.

d) All values of t so that the velocity is positive.

3. Hello, Dragon!

A particle moves along the x axis so that at time $\displaystyle t$ its position is given by: .$\displaystyle x(t)\:=\:\sin\left(\pi t^2\right)$

(A) Find the velocity at time $\displaystyle t.$
Velocity is the derivative of the position function.

. . $\displaystyle v(t) \;=\;x'(t) \;=\;2\pi t\cos(\pi t^2)$

(B) Find the acceleration at time $\displaystyle t.$
Acceleration is the derivative of the velocity function.

. . $\displaystyle a(t) \;=\;v'(t)\;=\;2\pi t\cdot\left[-2\pi t\sin(\pi t^2)\right] + 2\pi\cdot\cos(\pi t^2)$

. . . . . . $\displaystyle a(t) \;=\;-4\pi^2t^2\sin(\pi t^2) + 2\pi\cos(\pi t^2)$

(C) For what value of $\displaystyle t$ does the particle change direction?
Big hint: In order to change direction, the particle must come to a stop.
. .
The question becomes: When is the velocity equal to zero?

We have: .$\displaystyle 2\pi t\cos(\pi t^2) \:=\:0$

. . $\displaystyle (1)\;\;2\pi t \:=\:0 \quad\Rightarrow\quad\boxed{ t \:=\:0}$

. . $\displaystyle (2)\;\;\cos(\pi t^2) \:=\:0 \quad\Rightarrow\quad \pi t^2 \:=\:\frac{\pi}{2} + n\pi \quad\Rightarrow\quad t^2 \:=\:\frac{1}{2} + n \:=\:\frac{2n+1}{2}$

. . . . .$\displaystyle \boxed{t \;=\;\pm\sqrt{\frac{2n+1}{2}}}$ . for any integer $\displaystyle n \geq 0$

(D) All values of $\displaystyle t$ that the particle is moving to the left.
This part is very messy ... I'm still working on it.

If the particle is moving to the left, its velocity is negative.

We have: .$\displaystyle 2\pi t\cos(\pi t^2) \:<\:0$

. . Good luck!

4. Thank you so much, I found where i had made a tiny mistake. as for (d) i just grpaghed it and the answers i came up with was, the particle between -1 and 1 moves to the left from $\displaystyle \frac{-\sqrt2}{2}$ to $\displaystyle 0$ and from $\displaystyle \frac{\sqrt2}{2}$ to $\displaystyle 1$