The velocity time graph of a particle moving along a straight time as shown in figure. Calculate the displacement covered between t = 0 to t = 10 seconds.A 10 m B 100 m C 60 m D 20 m

A particle moves according to the equation; x = 10t2, where x is in meters and t is in seconds. Find the velocity for the time interval from 2.0 s to 2.1 s. (a) 0.1 m/s (b) 42 m/s (c) 44.1 m/s (d) 40 m/s (e) 2.0 m/s 11

The velocity function (in meters per second) is given for a particle moving along a line.v(t) = 3t − 8,    0 ≤ t ≤ 5(a) Find the displacement. m(b) Find the distance traveled by the particle during the given time interval. m

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

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 .

The displacement – time graph of a body in motion is given as shown in figure the velocity of the body is (in m/s)