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 $t=4$ ?

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Answer 1 · 2018-05-10T06:07:33

$(-43,22)$

Acceleration at $t = 4$ is

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

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

$a = (-43, 22)$

Direction:

$Tan θ = a_y/a_x = 22/(-43)$

$θ = Tan^-1(-22/43)$

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

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