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## Homework Statement

An object moves along the x axis according to the equation x = 3.00t

^{2}- 2.00t + 3.00 , where x is in meters and t is in seconds.

Determine :

(a) the average speed between t = 2.00 s and t = 3.00 s.

(b) the instantaneous speed at t = 2.00

(c) the average acceleration between t = 2.00 s and t = 3.00 s.

(d) the instantaneous acceleration at t = 2.00 s

(e) at what time is the object at rest?

## Homework Equations

average speed = [itex]\frac{d(total distance travelled)}{\Delta t}[/itex]

instantaneous speed = [itex]\frac{dx}{dt}[/itex]

[itex]\overline{a}[/itex] (average acceleration) = [itex]\frac{\Delta v}{\Delta t}[/itex]

a (instantaneous acceleration) = [itex]\frac{dv}{dt}[/itex] = [itex]\frac{dx^2}{dt}[/itex]

## The Attempt at a Solution

(a) I couldn't calculate the total distance.

(b) instantaneous speed = instantaneous velocity = [itex]\frac{dx}{dt}[/itex] = 6.00t - 2.00

instantaneous speed at t=2.00 = 6.00(2)-2.00 = 10 m/s

(c) [itex]\overline{a}[/itex] = [itex]\frac{\Delta v}{\Delta t}[/itex] = [itex]\frac{vf-vi}{1}[/itex] = [itex]\frac{v(3)-v(2)}{1}[/itex] = [itex]\frac{16-10}{1}[/itex] = 6 m/s

^{2}

(d) a = [itex]\frac{dx^2}{dt}[/itex] = 6 => a at 2.00s = 6 m/s

^{2}

(e) the rest means that v = 0

=> 6t-2=0

=>t=0.33 s //it seems wrong.