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 $t=1$ ?

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Answer 1 · 2017-10-27T05:49:35

Rate of accelaration is $3.606$ and direction is $-56.31^@$

Here velocity $v(t)=(t^2+2,cospit-3t)$ is given in parametric form,

accelaration is $a(t)=(dv(t))/(dt)=(d/(dt)(t^2+2),d/(dt)(cospit-3t))$

or $(2t,-pisinpit-3)$

and at $t=1$

$a(t)=(2,-3)$

This shows that $veca(t)=2hati-3hatj$

its magnitude is $sqrt(2^2+3^2)=sqrt(4+9)=sqrt(13)=3.606$

and direction is given by $tan^(-1)((-3)/2)=tan^(-1)(-1.5)=-56.31^@$

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=1 t=1 ?