A particle moves along a horizontal path, such that its velocity is given by v=3t2-6tm/s where t is the time in seconds. If it is initially located at the origin O, determine the distance travelled by the particle in time interval from t=0 to t=3.5 s and the particle's average velocity and average speed during the same time interval.

⇒  v=0  at   t=2 sFor t<2 s, velocity is negative. At t=2 s, velocity is zero and for t=2 s velocity is positive.s1=∫015vdt=∫01s3t2-6tdt =6.125 m = displacement upto 3.5 s s2=∫02vdt=∫023t2-6tdt =-4 m = displacement upto 2 s ∴  d= distance travelled in 3.5 s =4+4+6.125 =14.125 m  Average speed =dt=14.1253.5 =4.03 m/s  Average velocity =s1t=6.1253.5 =1.75 m/s.