An object's two dimensional velocity is given by $v(t) = (3t-sqrtt , t^3- 3t )$ . What is the object's rate and direction of acceleration at $t=5$ ?

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Answer 1 · 2017-04-27T06:47:48

The rate of acceleration is $=4ms^-2$ in the direction of $=73.6º$

Acceleration is the derivative of the velocity

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

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

$a(t)=v'(t)=(3-1/(2sqrtt),3t^2-3)$

When $t=2$ ,

$a(2)=(2.65,9)$

The rate of acceleration is

$||a(2)||=sqrt((2.65)^2+9^2)$

$=sqrt16=4$

The direction is

in the first quadrant

$theta=arctan(9/2.65)=73.6º$

An object's two dimensional velocity is given by v(t) = (3t-sqrtt , t^3- 3t )v(t) = (3t-sqrtt , t^3- 3t ). What is the object's rate and direction of acceleration at t=5 t=5 ?