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 ; `
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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.
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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.
`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.
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