English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

An object starts at time t = 0 with a velocity of v0 = +50 m/s and undergoes a constant acceleration of a = -11 m/s2.

I've already figured out a and b but I need help with c,d and e.

a)At what time does the speed of the object reach zero?
answer=4.56

b) How far from its starting (t = 0) position is the object at time t1?
answer=113.6

c) At what time does the object again pass through the starting (t = 0) position?

t2=____?___

d) If the object had initially been moving twice as fast (100 m/s), how far would it have gone before its velocity reached zero?
D=______?____

e) Suppose a second object begins moving with a constant speed of v = 50 m/s in the same direction from the same location at the same time as the object in part (d). At what time do the paths of these two objects once again cross?

tmeet= _____?____

2007-03-29 14:54:30 · 2 answers · asked by fabulous101 2 in Science & Mathematics Physics

2 answers

The equation of motion is
v(t)=v0-a*t
s(t)=s0+v0*t-.5*a*t^2

For this specific case
v(t)=50-11*t
s(t)=50*t-.5*11*t^2

a)
0=50-11*t
t=50/11
4.55

b
s(t)=50*4.55-.5*11*4.55^2
113.6

c
0=50*t-.5*11*t^2
t=50*2/11
=9.09

d
v(t)=100-11*t
t=100/11
=9.09

e
s1=100*t-.5*11*t^2
s2=50*t-.5*11*t^2

The question asks when is s1=s2 (They are at the same location)

Obviously they start at the same location
t=0

They never cross again since the slower one reaches apogee first and acclelerates negative at the same negative accleration as the second one that reaches apogee later.

j

2007-03-30 06:37:52 · answer #1 · answered by odu83 7 · 0 0

D = 55ft. t = sqrt(2D/g) t = sqrt(110ft/32ft/s^two) t = one million.86s I haven't any proposal the way you obtained zero.33s. Which if you happen to paintings it out backwards and also you remedy g. g = 2D/t^two g = 1010ft/s^two Acceleration = Velocity / Time Velocity = Distance / Time Essentially Acceleration = (Distance / Time) / Time = Distance / (Time)^two aka Distance over seconds squared. So no, you are not able to forget about the seconds squared while fixing for time that is squared. So you need to rectangular root the time to be able to get the precise importance of time, and no longer time squared.

2016-09-05 21:38:02 · answer #2 · answered by ? 4 · 0 0

fedest.com, questions and answers