Asked by Anonymous
The speed of a truck going south is reduced from 80 km/h to 44 km/h in a distance of 250 m. Calculate the following:
a. the time elapsed
b. the acceleration
c. the distance it traveled before it is brought to rest from 44 km/h
a. the time elapsed
b. the acceleration
c. the distance it traveled before it is brought to rest from 44 km/h
Answers
Answered by
Henry
Vo = 80 km/h * 1000 m/km * 1h/3600s. =
22.22 m/s.
Vf = (44/80) * 22.22 = 12.22 m/s.
b. t = (Vf - Vo) / a,
t = (12.22 - 22.22) / -0.689 = 14.51 s.
a. a = (Vf^2 - Vo^2) / 2d,
a=((12.22)^2 - (22.22)^2) / 500 =
- 0.689 m/s^2.
c. d = (Vf^2 - Vo^3) / 2a,
d = (0 - (12.22)^2) / -1.378 = 108.4 m.
22.22 m/s.
Vf = (44/80) * 22.22 = 12.22 m/s.
b. t = (Vf - Vo) / a,
t = (12.22 - 22.22) / -0.689 = 14.51 s.
a. a = (Vf^2 - Vo^2) / 2d,
a=((12.22)^2 - (22.22)^2) / 500 =
- 0.689 m/s^2.
c. d = (Vf^2 - Vo^3) / 2a,
d = (0 - (12.22)^2) / -1.378 = 108.4 m.
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