# Distance in Respect to Time

• Mar 19th 2009, 12:11 PM
Dragon
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
• Mar 19th 2009, 02:10 PM
Prove It
Quote:

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.
• Mar 19th 2009, 02:44 PM
Soroban
Hello, Dragon!

Quote:

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.

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

Quote:

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

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

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

Quote:

(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 $\displaystyle n \geq 0$

Quote:

(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!

• Mar 19th 2009, 06:15 PM
OnMyWayToBeAMathProffesor
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 $\displaystyle 1$