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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