# An object's two dimensional velocity is given by v(t) = ( 3t^2 - 2t , t ). What is the object's rate and direction of acceleration at t=1 ?

Aug 16, 2017

$a = 4.12$ ${\text{LT}}^{-} 2$

$\theta = {14.0}^{\text{o}}$

#### Explanation:

We're asked to find the magnitude and direction of the acceleration of an object at $t = 1$, given the velocity component equations.

Acceleration is the first derivative of velocity, so we differentiate the velocity component equations:

a_x(t) = d/(dt) [3t^2 - 2t] = ul(6t - 2

a_y(t) = d/(dt) [t] = ul(1

Substituting in $t = 1$:

${a}_{x} = 6 \left(1\right) - 2 = 4$

${a}_{y} = 1$

The magnitude of the acceleration is given by

a = sqrt((a_x)^2 + (a_y)^2) = sqrt(4^2 + 1^2) = color(red)(ulbar(|stackrel(" ")(" "4.12color(white)(l)"LT"^-2" ")|)

The ${\text{LT}}^{-} 2$ is the dimensional form of the units for acceleration. I used it here because no units were given.

The direction is given by

theta = arctan((a_y)/(a_x)) = arctan(1/4) = color(red)(ulbar(|stackrel(" ")(" "14.0^"o"" ")|)

measured anticlockwise from the positive $x$-axis (which is the normal measurement standard for planar angles).