The velocity function (in meters per second) is given for a particle moving along a line.v(t) = 3t − 8, 0 ≤ t ≤ 3(a) Find the displacement.(b) Find the distance traveled by the particle during the given time interval.

Answer:a) - 10.5 mb) 10.83 mStep-by-step explanation:Given:Velocity function , v(t) = 3t - 8,      0 ≤ t ≤ 3a) For displacement, integerating the function form time 0 to 3thus,[tex]\int\limits^3_0{v(t)} \, dt[/tex] = [tex]\int\limits^3_0 {3t - 8} \, dt[/tex]orDisplacement  =[tex][\frac{3}{2}t^2-8t]_0^3[/tex]orDisplacement  = [tex][\frac{3}{2}(3)^2-8(3)]-0[/tex]= - 10.5 mb) For total distance, let us first find the intervals where the velocity went from positive to negativethus,3t - 8 = 0t = [tex]\frac{8}{3}[/tex]the velocity changed it directionthus,we have the interval for speed as t ∈ [tex][0, \frac{8}{3}]\ to\ [\frac{8}{3},0][/tex]therefore,total distance =  [tex]\int\limits^{\frac{8}{3}}_0 {3t - 8} \, dt + \int\limits^3_{\frac{8}{3}} {3t - 8} \, dt[/tex][/tex]= [tex][\frac{3}{2}t^2-8t]_0^{\frac{8}{3}} +[\frac{3}{2}t^2-8t]^3_{\frac{8}{3}}[/tex]=  [tex][\frac{3}{2}(\frac{8}{3})^2-8(\frac{8}{3}) - (0)] + [\frac{3}{2}(3)^2-8(3)] - [\frac{3}{2}(\frac{8}{3})^2-8(\frac{8}{3})][/tex]= [tex]\frac{32}{3}-\frac{21}{2}+\frac{32}{3}[/tex]= [tex]\frac{65}{6}[/tex]= 10.83 m