First, we find the time spent accelerating:
s = ut + 1/2 at²; u = 0
10 = 1/2 5t²
t = 2 seconds
Velocity at point B:
v = u + at; u = 0, a = 5 m/s², t = 2 s
v = (5)(2) = 10 m/s
Now, the same velocity is at C so we find the time decelerating to 0 in CD
v = u + at, v = 0, u = 10 m/s, a = -4 m/s²
0 = 10 - 4t
t = 2.5 seconds
Total time = time(AB) + time(BC) + time(CD)
20 = 2 + time(BC) + 2.5
time(BC) = 15.5 seconds
Distance = velocity x time
BC = 10 x 15.5
= 155 m
Answer:
£66.4
Step-by-step explanation:
£332/5
Answer:
ummm I think u want the answer x= 11 1/4
Volume = length x Base x height
3240 = 18 x 12 x H
3240 = 216 x H
H = 3240 / 216 = 15 inches
3a+2.25=3a+1.75
Implies 3a+2.25=3a+1.75+.5
Answer is 0.5