At time t=0, a car moving along a straight line has a velocity of 16 ms-1. It slows down with an acceleration of -0.5t ms-2, where t is in second. Mark the correct statement (s).

a=-0.5t=dvdt∴ ∫16vdv=∫0t-0.5tdt∴ v=16-0.25t2s=∫vdt=16t=0.25t33v=0 when 16-0.25t1=0or t=8 sSo direction of velocity changes at 8s. Up to 8s distance = displacement∴ At 4sd=16×4-0.25×(4)33=58.67 mS8s=16×8-(0.25)(8)33=85.33 mS10s=16×10-(0.25)(10)33 =76.67 cmDistance travelled in 10 s,d =(85.33)+(85.33-76.67) =94 mAt 10 s,  v=16-0.25(10)2=-9 m/s∴ Speed =9 m/s