A particle moves with acceleration `a(t)` m/s^2 along an s-axis and has velocity `v_0` m/s at time `t=0` . Find the displacement AND the distance traveled by the particle during the given time interval.
`a(t)=9 ; `
`v_0=-1 ; `
2 Answers | Add Yours
`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.
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.
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.
We’ve answered 328,308 questions. We can answer yours, too.Ask a question