An object's two dimensional velocity is given by $v (t) = (t^{2} + 2 t, cos π t - 3 t)$ $v(t) = ( t^2 +2t , cospit - 3t )$ . What is the object's rate and direction of acceleration at $t = 1$ $t=1$ ?

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An object's two dimensional velocity is given by $v (t) = (t^{2} + 2 t, cos π t - 3 t)$ $v(t) = ( t^2 +2t , cospit - 3t )$ . What is the object's rate and direction of acceleration at $t = 1$ $t=1$ ?

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Answer 1 · 2017-12-25T19:44:27

$a = 5$ $a = 5$
$θ = - 37^{o} or 37^{o}$ $theta = - 37^o " or " 37"^o$ SE

Explanation:

$a (t) = \frac{d v (t)}{d t} = \frac{d}{d t} (t^{2} + 2 t, cos π t - 3 t) = (2 t + 2, - π sin π t - 3)$ $a(t) =(dv(t))/(dt) =d/(dt)(t^2+2t, cos pit - 3t) = ( 2t+2, -pi sin pit-3)$

a(1) = (4, -3)
The magnitude of the acceleration is
$| | a (1) | | = \sqrt{4^{2} + {(- 3)}^{2}} = 5$ $||a(1)|| = sqrt(4^2 + (-3)^2)= 5$

The direction of the acceleration is determined from

$tan θ = \frac{v_{y}}{v_{x}}$ $tantheta = (v_y)/(v_x)$
$θ = {tan}^{- 1} (\frac{v_{y}}{v_{x}}) = {tan}^{- 1} (- \frac{3}{4}) = - 37^{o}$ $theta = tan^-1((v_y)/(v_x) )= tan^-1(-3/4)=-37^o$
or

$θ = 37^{o}$ $theta = 37^o$ SE

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An object's two dimensional velocity is given by $v (t) = (t^{2} + 2 t, cos π t - 3 t)$ $v(t) = ( t^2 +2t , cospit - 3t )$ . What is the object's rate and direction of acceleration at $t = 1$ $t=1$ ?

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An object's two dimensional velocity is given by v(t) = ( t^2 +2t , cospit - 3t )v(t)=(t2+2t,cosπt−3t)v(t) = ( t^2 +2t , cospit - 3t ). What is the object's rate and direction of acceleration at t=1 t=1t=1 ?

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Explanation:

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An object's two dimensional velocity is given by $v (t) = (t^{2} + 2 t, cos π t - 3 t)$ $v(t) = ( t^2 +2t , cospit - 3t )$ . What is the object's rate and direction of acceleration at $t = 1$ $t=1$ ?