A box slides down an incline with uniform acceleration. It starts from rest and attains a speed of 4.8 m/sec in 6sec. Find the distance between t=3 and t=6.

We have u=0 m/s, v=4.8 m/s, t=6 secSo,》a=(v-u)/t=4.8/6=0.8 m/s^2Distance in 3 sec 》S3=ut+1/2at^2 》S3=0+1/2(0.8×3×3) 》S3=3.6mAnd distance in 6 sec》S6=ut+1/2at^2》S6=0+1/2(0.8×6×6)》S6=14.4m So distance moved between 3 to 6 sec》S=14.4-3.6》S=10.8m Ans..If you like this approve my answer please.