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

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

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Answer 1 · 2018-05-10T06:07:33

$(- 43, 22)$ $(-43,22)$

Explanation:

Acceleration at $t = 4$ $t = 4$ is

$a_{x} (4) = \frac{d}{d t} (v_{x}) = 5 - 3 t^{2} = 5 - (3 \times 4^{2}) = 5 - 48 = - 43$ $a_x(4) = d/dt(v_x) = 5 - 3t^2 = 5 - (3 × 4^2) = 5 - 48 = -43$

$a_{y} (4) = \frac{d}{d t} (v_{y}) = 6 t - 2 = (6 \times 4) - 2 = 22$ $a_y(4) = d/dt(v_y) = 6t - 2 = (6 × 4) - 2 = 22$

$a = (- 43, 22)$ $a = (-43, 22)$

Direction:

$Tan θ = a_y/a_x = 22/(-43)$

$θ = Tan^-1(-22/43)$

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

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

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An object's two dimensional velocity is given by v(t) = ( 5t-t^3 , 3t^2-2t)v(t)=(5t−t3,3t2−2t)v(t) = ( 5t-t^3 , 3t^2-2t). What is the object's rate and direction of acceleration at t=4 t=4t=4 ?

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Explanation:

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