Better Students Ask More Questions.
A particle moves with acceleration `a(t)` m/s^2 along an s-axis and has velocity...
2 Answers | add yours
The particle moves along a straight line with initial velocity `v_0` = -1 m/s and acceleration a(t) = 9 m/s^2. The distance traveled by the particle during a duration of time equal to t is `s = v_0*t + (1/2)*a(t)*t^2` .
Substituting the given values the displacement of the particle in the period from t = 0 to t = 2 is:
(2 - 0)*9 + (1/2)*9*(2 - 0)
=> 18 + 9
The particle travels 27 m in the given interval.
Posted by justaguide on August 12, 2013 at 5:40 PM (Answer #1)
1 Reply | Hide Replies ▲
Thank you for your response. Wiley Plus is saying 27m is incorrect. I should have mentioned it, but we are asked to solve using integration, where:
and distance traveled = `int_(t_0)^(t_1)|v(t)|dt`
I found that displacement = 16, but am not sure where I went wrong trying to obtain distance traveled (it does not equal 16, atleast according to wiley) Thank you in advance if you are able to help clarify.
Posted by user7230927 on August 12, 2013 at 5:50 PM (Reply #1)
`a(t) = (dV(t))/(dt)`
`dV(t) = a(t)dt`
`intdV(t) = inta(t)dt`
`V(t) = int9dt`
`V(t) = 9t+C`
`V(0) = 9xx0+C`
`-1 = C`
`V(t) = 9t-1`
`V(t) = (dS(t))/(dt)`
`S(t) = intV(t)dt`
`S(t) = int(9t-1)dt`
`S(t) = 9t^2/2-t+D` where D is a constant.
It is given that the time interval is `0<=t<=2`
Displacement of particle `= S(2)-S(0)`
Displacement of particle `= (9xx4/2-2) = 16m`
For distance travelled;
`S(t) = int_0^2(9t-1)dt`
`S(t) = [9t^2/2-t]_0^2`
`S(t) = (9xx4/2-2) = 16`
So the displacement and distance travelled by the particle is 16m.
Posted by jeew-m on August 12, 2013 at 7:10 PM (Answer #2)
Join to answer this question
Join a community of thousands of dedicated teachers and students.