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

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Answer 1 · 2017-01-25T04:35:18

Magnitude is $15.8115$ and direction is $(-18.435^@)$

As $v(t)=(t^2+2t,2cospit-t)$ ,

at $t=3$ , $v(3)=(3^2+2xx3,2cos3pi-3)$

= $(9+6,2xx(-1)-3)$ = $(15,-5)$

Hence object's rate is $|v(3)|=sqrt(15^2+5^2)$

= $sqrt250=5sqrt10=5xx3.1623=15.8115$

and direction is given by $tan^(-1)((-5)/15)=tan^(-1)((-1)/3)=-18.435^@$
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An object's two dimensional velocity is given by v(t) = ( t^2 +2t , 2cospit - t )v(t) = ( t^2 +2t , 2cospit - t ). What is the object's rate and direction of acceleration at t=3 t=3 ?