Recent content by alsy

m/s2

ok a= -1.6 m/s

its correct now a=1.6 m/s

0=4.8+3a a=1.6 m/s

u=initial velocity a=acceleration t=time 8-4=4 constant velocity

0= 4.8+ a(3) 0=4.8 +3a a=-1.6 m/s

0= 4.8+ a(4) 0= 4.8 +4A a=-1.2 m/s

v=u+at I have calculated part one as v=u+at 0+1.2(4) =4.8 m/s the part 2 deceleration I'm using same formula n v=u+at and as you say using constant velocity of 4 4+(1.2)(4)=8.8 m/s that's what I'm getting

4+(1.2)(4)=8.8 m/s

v=u+at 8+(1.2)(4)=12.8 m/s

v=u+at 0+1.2(4) =4.8 m/s

ok deceleration v = u +at

units would be m/s

v = u +at I have no deceleration formula

Homework Statement a lift accelerates from rest at 1.2m/s2 for 4 seconds..it travels at a constant velocity for 8 seconds and comes to rest in 3 seconds.calculate 1)the maximum velocity 2)deceleration in the final 3 secondsHomework Equations calculate 1)the maximum velocity 2)deceleration in...