A box slides down an incline with uniform acceleration. it starts at 0m/s and reaches 2.7m/s in 3s.
Find the acceleration and the distance moved int he first 6s.
2007-08-15 16:23:37 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
when solving this equation, the angle of inclination is needed because v2=u2+2as u=0 v=2.7 acc=gsin(angle of inclination) where g in 10m/s2 and t=3s
so when the angle of inclination in known,using formula above,the acceleration and distance travelled can be calculated.I hope you understand what i mean.
2007-08-16 03:17:41 · answer #1 · answered by Emperor 3 · 0⤊ 0⤋
a = v' - v / t
given that v' = 2.7 m/s, v = 0 m/s, t = 3 s
thus
a = 2.7 / 3 = 0.9 m/s^2
now d = 1/2*a*t^2
at t = 6 s
d = 1/2*0.9*6^2
d = 16.2 m
2007-08-15 23:30:05 · answer #2 · answered by quigonjan 3 · 0⤊ 0⤋
start with v = at where t = time
a = v/t = 2.7m/s/3s = 0.9 m/s^2
distance = 1/2at^2 = 1/2*0.9*36 = 16.2 m
2007-08-15 23:27:54 · answer #3 · answered by nyphdinmd 7 · 0⤊ 0⤋