# The motion of the particle is given parametrically by x(t)=3t^2-3, y(t)=2t +2 for t is greater than or equal to 0, how do you find the speed of the particle at t=1?

Jun 26, 2016

$2 \sqrt{10}$

#### Explanation:

$x = 3 {t}^{2} - 3 , \dot{x} = 6 t$
$y = 2 {t}_{2} , \dot{y} = 2$

where $\dot{x} , \dot{y}$ are the velocities in the x and y directions

speed s is the magnitude of the velocity vector and so $s = \sqrt{{\dot{x}}^{2} + {\dot{y}}^{2}}$

at $t = 1 , \dot{x} = 6 , \dot{y} = 2$ so

$s = \sqrt{40} = 2 \sqrt{10}$