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

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

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Answer 1 · 2017-01-31T12:06:16

$a (t) = d \frac{v}{d t} = (2 t, - π sin π t - 3)$ $a(t) = dv/dt = (2t, -pi sin pit - 3)$ at t=2 one finds
$a (2) = (4, - 3)$ $a(2) = (4,-3)$ The magnitude f the acceleration vector is
$\sqrt{16 + 9} = 5$ $sqrt(16+9) = 5$
Its direction is given by an angle theta such that $tan θ = - \frac{3}{4}$ $tan theta = -3/4$ or $θ = - 37$ $theta = -37$ degrees.

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

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

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