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

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Answer 1 · 2017-05-06T08:35:55

$"answer :"a(1)=(0,-1)$

$d/(d t) v(t)=(color(red)(d/(d t)(t^2-2t)),color(green)(d/(d t)(cos pi t-t)))$

$d/(d t) v(t)=a(t)$

$a(t)=(color(red)(2t-2),color(green)(-pi sin pi t-1))$

$"solve t=1"$

$a(1)=(2*1-2,-pi sin pi *1-1)$

$a(1)=(0,-pi sin pi-1)$

$sin pi=0$

$a(1)=(0,-1)$

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