a particle moving in a straight line with constant acceleration has a velocity of 4 ms-1 at one instant, and 6 seconds later, it has a velocity of 22 ms-1. find the acceleration, and the distance covered by the particle in the 6 seconds.

Using concepts of Motion, we got 78m is the distance covered by the particle in the 6 seconds.We are given that particle is moving with constant acceleration, it means we can apply three laws of motion to find the solution.Firstly, we need to find the acceleration,according to first law of motion,v=u + atwhere u is initial velocity, v is final velocity, a is constant acceleration and t tis the timeTherefore we are given final velocity(v)=22m/s, initial velocity(u)=4m/s and time=6 secTherefore on putting the value in above formula,=>22 = 4 +a×6=>6×a=22-4=>6a=18=>a=3m/s² So constant acceleration =3m/s²Now we need to find the distance covered by the particle in 6 secondsSo, we apply second law of motion S= ut + (1/2)at²where S is displacement/distance, u is initial velocity, a is acceleration and t is the time=>S=(4×6)+[(1/2)×3×6×6]=>S=24+54=>S=78mHence, distance covered by the particle in the 6 sec is 78mTo know more about Motion, visit here:https://brainly.com/question/22810476#SPJ4