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

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

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Answer 1 · 2017-01-31T12:06:16

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

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

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