Motion in one dimension problem

  • Thread starter Ammar w
  • Start date
  • #1
28
0

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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
668
0
Looks like you did pretty well.

The total distance? x= distance,
so calculate x at 3 secs minus x at 2 secs
 
  • #3
28
0
thanks
but x at 3 minus x at 2 is the displacement which we use to calculate the average velocity not average speed. right??
 
  • #4
668
0
Speed is just (absolute) velocity.
 
  • #5
28
0
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)
 
  • #6
668
0
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.
 
  • #7
28
0
sorry, still not convinced!
 

Related Threads on Motion in one dimension problem

  • Last Post
Replies
1
Views
3K
Replies
1
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
960
  • Last Post
Replies
3
Views
954
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
10
Views
207
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
4K
Top