# An object's two dimensional velocity is given by v(t) = ( 5t-t^3 , 3t^2-2t). What is the object's rate and direction of acceleration at t=4 ?

May 10, 2018

$\left(- 43 , 22\right)$

#### Explanation:

Acceleration at $t = 4$ is

a_x(4) = d/dt(v_x) = 5 - 3t^2 = 5 - (3 × 4^2) = 5 - 48 = -43

a_y(4) = d/dt(v_y) = 6t - 2 = (6 × 4) - 2 = 22

$a = \left(- 43 , 22\right)$

Direction:

Tan θ = a_y/a_x = 22/(-43)

θ = Tan^-1(-22/43)

Where $\theta$ is angle with $x$-axis