An object's two dimensional velocity is given by $v (t) = (3 t^{2} - 2 t, 1 - 3 t)$ $v(t) = ( 3t^2 - 2t , 1- 3t )$ . What is the object's rate and direction of acceleration at $t = 4$ $t=4$ ?

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Answer 1 · 2017-05-10T06:52:27

The rate of acceleration is $= 22.2 m s^{- 1}$ $=22.2ms^-1$ in the direction of $=352.2º$

The acceleration is the derivative of the velocity.

$v(t)=(3t^2-2t, 1-3t)$

$a(t)=v'(t)=(6t-2, -3)$

Therefore,

$a(4)=(24-2, -3)=(22,-3)$

The rate of acceleration is

$||a(4)||=sqrt(22^2+(-3)^2)$

$=sqrt493$

$=22.2ms^-2$

The direction is

$theta=arctan(-3/22)$

$theta$ lies in the 4th quadrant

$theta=360-7.8=352.2º$

An object's two dimensional velocity is given by v(t) = ( 3t^2 - 2t , 1- 3t )v(t)=(3t2−2t,1−3t)v(t) = ( 3t^2 - 2t , 1- 3t ). What is the object's rate and direction of acceleration at t=4 t=4t=4 ?