An object's two dimensional velocity is given by $v(t) = ( t-2t^3 , 4t^2-t)$ . What is the object's rate and direction of acceleration at $t=2$ ?

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Answer 1 · 2017-04-26T05:45:46

The rate of acceleration is $=27.5$ in the direction of $=146.9º$

Acceleration is the derivative of the velocity

$a(t)=v'(t)$

$v(t)=(t-2t^3,4t^2-t)$

$a(t)=v'(t)=(1-6t^2,8t-1)$

When $t=2$ ,

$a(2)=(-23,15)$

The rate of acceleration is

$||a(2)||=sqrt((-23)^2+15^2)$

$=sqrt754=27.5$

The direction is

in the second quadrant

$theta=180-arctan(15/23)=146.9º$

An object's two dimensional velocity is given by v(t) = ( t-2t^3 , 4t^2-t)v(t) = ( t-2t^3 , 4t^2-t). What is the object's rate and direction of acceleration at t=2 t=2 ?