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:
5 4/9 - 2 5/9= 26/9 or 2 8/9
Step-by-step explanation:
