The relation 3𝑡=3𝑥+6  describes the displacement of a particle in one direction where 𝑥  is in metres and t in sec. The displacement, when velocity is zero, is

The displacement of a particle moving in a straight line is described by the relation 𝑆=6+12𝑡−2𝑡2. Here S in meter and t is in sec. The distance covered by the particle in first 5s is

The displacement of a particle starting from rest a t = 0 is given byS=6t2–t3. The time in seconds at which the particle will attain zero velocity is  (a) 8s (b) 6s (c) 4s (d) 3s

𝑎(𝑡)=8.7𝑡2+6,where 𝑡 is measured in seconds.1. Write an equation for the velocity 𝑣 at time 𝑡. Simplify your answer.𝑣(𝑡)=

A particle moves along a straight line such that its displacement at any time t is given by S = t3 – 6t2 + 3t + 4 metres. The velocity when the acceleration is zero is :4 ms–1– 12 ms–1 42 ms–1– 9 ms–1

EXAMPLE 6 A particle moves along a line so that its velocity at time t is v(t) = t2 − t − 20 (measured in meters per second).(a) Find the displacement of the particle during 1 ≤ t ≤ 6.(b) Find the distance traveled during this time period.SOLUTION(a) By this equation, the displacement iss(6) − s(1)  =  6v(t) dt1 =  6(t2 − t − 20) dt1 =  t33​ − t22 − 20t 61 =  −2756​ .This means that the particle moved approximately 45.83 meters to the left.(b) Note that v(t) = t2 − t − 20 = (t − 5)(t + 4) and so v(t) 0 on the interval [1, 5] and v(t) 0 on [5, 6]. Thus, from this equation, the distance traveled is6|v(t)| dt1 =  5[−v(t)] dt1 + 6v(t) dt5 =  5(−t2 + t + 20) dt1 + 6(t2 − t − 20) dt5 =  −t33​+t22​+20t 51 +  t55​+t22​+20t 65 =  9.17 .