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 · 2017-06-16T08:06:11

The rate of acceleration is $=3.47ms^-2$ in the direction $=196.8º$ anticlockwise from the x-axis

The acceleration is the derivative of the velocity

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

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

So,

when $t=8$

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

$=(-3.32,-1)$

The rate of acceleration is

$=||a(3)||=sqrt(3.32^2+(1)^2)=sqrt12.04=3.47$

The direction is $theta=180+arctan(1/3.32)=196.8º$

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 ?