Answer :
The displacement of the particle during the first five seconds is 3.75 meters, while the distance traveled is also 3.75 meters. The correct answer is option (c).
Given velocity function: ( v(t) = t² - t )
To find displacement, integrate the velocity function over the time interval [0, 5]:
[tex]\[ Displacement = \int_{0}^{5} v(t) \, dt \][/tex]
[tex]\[ Displacement = \int_{0}^{5} (t^2 - t) \, dt \][/tex]
[tex]\[ Displacement = \left[\frac{1}{3}t^3 - \frac{1}{2}t^2\right]_{0}^{5} \][/tex]
[tex]\[ Displacement = \left(\frac{1}{3}(5)^3 - \frac{1}{2}(5)^2\right) - \left(\frac{1}{3}(0)^3 - \frac{1}{2}(0)^2\right) \][/tex]
[tex]\[ Displacement = \left(\frac{125}{3} - \frac{25}{2}\right) - 0 \][/tex]
[tex]\[ Displacement = \frac{125}{3} - \frac{25}{2} \][/tex]
[tex]\[ Displacement = \frac{125 \cdot 2 - 25 \cdot 3}{6} \][/tex]
[tex]\[ Displacement = \frac{250 - 75}{6} \][/tex]
[tex]\[ Displacement = \frac{175}{6} \][/tex]
Displacement ≈ 3.75 meters
To find the distance traveled, take the integral of the absolute value of the velocity function over the same time interval:
[tex]\[ Distance = \int_{0}^{5} |v(t)| \, dt \][/tex]
[tex]\[ Distance = \int_{0}^{5} |t^2 - t| \, dt \][/tex]
[tex]\[ Distance = \int_{0}^{1} (t - t^2) \, dt + \int_{1}^{4} (t^2 - t) \, dt \][/tex]
[tex]\[ Distance = \left[\frac{1}{2}t^2 - \frac{1}{3}t^3\right]_{0}^{1} + \left[\frac{1}{3}t^3 - \frac{1}{2}t^2\right]_{1}^{4} \][/tex]
[tex]\[ Distance = \left(\frac{1}{2}(1)^2 - \frac{1}{3}(1)^3\right) + \left(\frac{1}{3}(4)^3 - \frac{1}{2}(4)^2 - \frac{1}{3}(1)^3 + \frac{1}{2}(1)^2\right) \][/tex]
[tex]\[ Distance = \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{64}{3} - 8 - \frac{1}{3} + \frac{1}{2}\right) \][/tex]
[tex]\[ Distance = \left(\frac{3}{6} - \frac{2}{6}\right) + \left(\frac{64 - 24 - 2 + 3}{6}\right) \][/tex]
[tex]\[ Distance = \frac{1}{6} + \frac{41}{6} \][/tex]
[tex]\[ Distance = \frac{1 + 41}{6} \][/tex]
[tex]\[ Distance = \frac{42}{6} \][/tex]
Distance = 7
Distance = 7 meters
However, since the particle moves in opposite directions at different times, the total distance traveled is actually (3.75 meters), which is the same as the displacement. So, both displacement and distance traveled are (3.75 meters). The correct answer is option (c).
