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

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Answer 1 · 2017-05-20T12:44:40

The rate of acceleration is $=10.05ms^-2$ and the direction is $=5.7º$ anticlockwise from the x-axis

The acceleration is the derivative of the velocity.

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

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

Therefore,

$a(2)=(10, 1)$

The rate of acceleration is

$||a(2)||=sqrt(10^2+1)$

$=sqrt101$

$=10.05ms^-2$

The direction is

$theta=arctan(1/10))$

$theta$ lies in the 1st quadrant

$theta=5.7º$ anticlockwise from the x-axis

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