Motion in one dimension problem

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    Dimension Motion
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Homework Help Overview

The problem involves an object moving along the x-axis described by a quadratic equation. Participants are tasked with calculating average and instantaneous speeds and accelerations at specific times, as well as determining when the object comes to rest.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of total distance versus displacement in relation to average speed and average velocity. There is a focus on the definitions and distinctions between speed and velocity.

Discussion Status

Some participants have provided guidance on calculating average speed and have engaged in clarifying the differences between average speed and average velocity. There is ongoing debate regarding the interpretation of distance and displacement in the context of the problem.

Contextual Notes

Participants are exploring the implications of the object's motion as described by the equation, with specific attention to the increasing nature of position x over time. There is uncertainty about the definitions being applied and the calculations involved.

Ammar w
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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 = [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/s2

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