# Math Help - Magnitue

1. ## Magnitue

I was doing just fine on my homework until I came across this bad boy. The x and y coordinates of a moving particle are given by the parametric equations $x = t^3 = 5t$ and $y = 4t^2 - 3t^3$ where $t$ is in seconds, and x and y are in metres. To 3 significant digits, find the magnitute and direction of the acceleration at $t=1\; s$.

2. Originally Posted by R3ap3r
I was doing just fine on my homework until I came across this bad boy. The x and y coordinates of a moving particle are given by the parametric equations $x = t^3 = 5t$ and $y = 4t^2 - 3t^3$ where $t$ is in seconds, and x and y are in metres. To 3 significant digits, find the magnitute and direction of the acceleration at $t=1\; s$.
$r(t)=(t^3+5t) \vec i + (4t^2-3t^3) \vec j$

taking the derivative

$v(t)=(3t^2+5) \vec i + (8t-9t^2) \vec j$

one more time

$a(t)=(6t) \vec i + (8-18t) \vec j$

The direction is given by a(1)

$a(1)=6 \vec i - 10 \vec j$

the magnitue is

$|a(1)|=\sqrt{(6)^2+(-10)^2}=\sqrt{136}=2\sqrt{34}$

3. Alright so the answer is 11.7 magnitue. How do I find the direction?

4. Originally Posted by R3ap3r
Alright so the answer is 11.7 magnitue. How do I find the direction?
$a(1)=6 \vec i - 10 \vec j$

this is the vector pointing in the direction of acceleration.

if you need the angle we take the arctan

$\theta=\tan^{-1}\left( \frac{y}{x}\right)=\tan^{-2}\left( \frac{-10}{6}\right) \approx -59.04^\circ$