MHB Mechanics- general motion in a straight line.

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A particle's velocity is defined by the equation v = -t^3 + 9t for the time interval 0 < t < 5 seconds. The displacement of the particle at t = 5 seconds is calculated to be -43.8 meters, indicating it has moved 43.8 meters in the negative direction from its original position. To find the total distance traveled from t = 0 to t = 5, the velocity is analyzed, showing it is positive from 0 to 3 seconds and negative from 3 to 5 seconds. The correct approach involves integrating the absolute value of velocity over these intervals, leading to a total distance of 84.3 meters. Understanding the distinction between displacement and distance is crucial in solving these types of motion problems.
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A particle moves in a straight line. The velocity of the particle, v m/s, at time t s is given by v= -t^3+9t m/s for 0<t<5
a) Find the displacement of the particle from its original position, when t=5s
I got the ans for this by integration and limits 5 and 0 =- 43.8
b) work out the distance that the particle travels from t= 0 to t=5
I don't understand this. Velocity is positive from 0 to 3 and negative from 3 to 5 when I plot the velocity time graph.
I tried integration again with limits 3 to 0 and the next limit from 5 to 3. Iam not getting the ans which is 84.3m
 
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in general, distance traveled is the integral of speed …

$\displaystyle D = \int_{t_0}^{t_f} |v(t)| \, dt$

Note the velocity in this problem is positive in the interval (0,3) and negative in the interval (3,5]

two ways to do this …

$\displaystyle D = \int_0^3 9t-t^3 \, dt + \int_3^5 t^3 - 9t \, dt$

$\displaystyle D = \int_0^3 9t-t^3 \, dt - \int_3^5 9t-t^3 \, dt$
 
skeeter said:
in general, distance traveled is the integral of speed …

$\displaystyle D = \int_{t_0}^{t_f} |v(t)| \, dt$

Note the velocity in this problem is positive in the interval (0,3) and negative in the interval (3,5]

two ways to do this …

$\displaystyle D = \int_0^3 9t-t^3 \, dt + \int_3^5 t^3 - 9t \, dt$

$\displaystyle D = \int_0^3 9t-t^3 \, dt - \int_3^5 9t-t^3 \, dt$
Thank you very much!
 
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