An object's two dimensional velocity is given by $v(t) = ( tsin(pi/3t) , 2cos(pi/2t )- t )$ . What is the object's rate and direction of acceleration at $t=8$ ?

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Answer 1 · 2017-06-01T12:36:53

The rate of acceleration is $=1.71ms^-2$ in the direction $35.7º$ clockwise from the x-axis.

The acceleration is the derivative of the velocity.

$v(t)=(tsin(1/3pit), 2cos(1/2pit)-t)$

$a(t)=v'(t)=(1*sin(1/3pit)+t*1/3picos(1/3pit), -2*1/2pisin(1/2pit)-1)$

$=(sin(1/3pit)+1/3pitcos(1/3pit),-pisin(1/2pit)-1)$

Therefore,

$a(8)=(sin(8/3pi)+8/3picos(8/3pi),-pisin(4pi)-1)$

$=(1.39,-1)$

The rate of acceleration is

$||a(8)||=sqrt(1.39^2+1^2)$

$=sqrt(2.93)$

$=1.71ms^-2$

The direction is

$theta=arctan(-1/1.39)=-35.7º$

An object's two dimensional velocity is given by v(t) = ( tsin(pi/3t) , 2cos(pi/2t )- t )v(t) = ( tsin(pi/3t) , 2cos(pi/2t )- t ). What is the object's rate and direction of acceleration at t=8 t=8 ?