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

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Answer 1 · 2017-01-25T07:42:00

Rate of acceleration is $1.0427$ and direction is $62.364^@$

As $v(t)=(sin(pi/3t),2cos(pi/2t)-t)$ ,

$a(t)=d/(dt)v(t)=(pi/3cos(pi/3t),-2xxpi/2xxsin(pi/2t)-1)$

at $t=2$ , $a(2)=(pi/3cos((2pi)/3),-pisin(pi)-1)=(-pi/6,-1)$

Hence object's rate of acceleration is $|a(2)|=sqrt((pi/6)^2+(-1)^2)$

= $sqrt(pi/36+1)=sqrt(1.0873)=1.0427$

and direction is given by $tan^(-1)((-1)/(-pi/6))=tan^(-1)(6/pi)=62.364^@$

An object's two dimensional velocity is given by v(t) = ( sin(pi/3t) , 2cos(pi/2t )- t )v(t) = ( sin(pi/3t) , 2cos(pi/2t )- t ). What is the object's rate and direction of acceleration at t=2 t=2 ?