The velocity function (in meters per second) is given for a particle moving along a line.

Consider the given velocity function: v(t)-3 -8, 0 <t<5 The objective is to find the displacement. The displacement function is obtained by integrating the velocity function Thus, displacement- |v(t (3t - 8)dt 3 0 52-0)-8(5-0 (25)-8(5) 75 40

75-80 2 -2.5 Thus, the displacement is -2.5 m The objective is to find the distance travelled by the particle during t- 0 to t- 5. The distance travelled is obtained by the integration of the absolute value of the velocity function Thus, we have to find the interval on which velocity is positive and on which it is negative To find critical point, put v(t) 0 3t-8-0

Thus, v(f) changes its sign at j if it does Pick any value in the interva0,syt 2 and check the sign of v(t) At t 2, v(t)3(2)-8 6-82<0 >v() is negative in the interval 0, Pick any value in the intervalsay 3 and check the sign of v(t) At t 3, v(t) 3(3)-8-9-8 1>0 v(t) is positive in the interval 0,

Hence, Distance -Jlv(r)dr 3 (8 0 +805

3 64 8 3 +8+25 3)2 15 -8 64 2 9 32 64 3(225-64 32 64 161 56 113 18.33 Thus, the distance travelled is 18.33m