Motion in one dimension problem

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The discussion focuses on a motion problem involving an object moving along the x-axis, described by the equation x = 3.00t² - 2.00t + 3.00. Participants calculate average speed, instantaneous speed, average acceleration, and instantaneous acceleration at specific times, with results indicating average speed between t = 2.00 s and t = 3.00 s is based on displacement, while instantaneous speed at t = 2.00 s is determined to be 10 m/s. There is a distinction made between average speed and average velocity, emphasizing that average speed is the total distance traveled divided by time, while average velocity considers displacement. The instantaneous acceleration at t = 2.00 s is calculated to be 6 m/s², and the object is determined to be at rest at t = 0.33 s. The discussion highlights confusion over definitions of speed and velocity, clarifying their differences in the context of the problem.
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Homework Statement


An object moves along the x-axis according to the equation x = 3.00t2 - 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 = \frac{d(total distance travelled)}{\Delta t}

instantaneous speed = \frac{dx}{dt}

\overline{a} (average acceleration) = \frac{\Delta v}{\Delta t}

a (instantaneous acceleration) = \frac{dv}{dt} = \frac{dx^2}{dt}

The Attempt at a Solution



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

(b) instantaneous speed = instantaneous velocity = \frac{dx}{dt} = 6.00t - 2.00
instantaneous speed at t=2.00 = 6.00(2)-2.00 = 10 m/s

(c) \overline{a} = \frac{\Delta v}{\Delta t} = \frac{vf-vi}{1} = \frac{v(3)-v(2)}{1} = \frac{16-10}{1} = 6 m/s2

(d) a = \frac{dx^2}{dt} = 6 => a at 2.00s = 6 m/s2

(e) the rest means that v = 0
=> 6t-2=0
=>t=0.33 s //it seems wrong.
 
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Looks like you did pretty well.

The total distance? x= distance,
so calculate x at 3 secs minus x at 2 secs
 
thanks
but x at 3 minus x at 2 is the displacement which we use to calculate the average velocity not average speed. right??
 
Speed is just (absolute) velocity.
 
I think no.\
the average velocity is different from average speed
average velocity = dispacement/time
average speed = distance/time
sometimes average velocity is zero (when a particle moves and return to the starting point) where the average speed is a positive number.

in this problem I think the distance is like what you said = x at 3s - x at 2s because the position x is always increasing (from the equation)
 
Speed is defined as the scalar magnitude of the velocity vector. Since this is a one dimensional problem it simply becomes the absolute velocity.
Fact.
 
sorry, still not convinced!
 
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