a = ∆v/∆t
distance = d = 1/2 at^2
F = ma
work = w = Fd
P = w/t
So plug in your numbers
A 2200 kg car can accelerate from 0 to 30.0 m/s in 5.5 seconds. What is the power required?
4 answers
I have to get better about reading the question and figuring out what I need to know, then determining what I need to find in order to find the solution. I just look at it and panic and think I don't have all of the information to do it.
thanks!
a= 30/5.5=5.45~5.5
d=1/2(5.5)(5.5)^2=83m
w f=2200(5.5)=12100 x 83=1004300
p= 1004300/5.5
Answer: 182600 W
thanks!
a= 30/5.5=5.45~5.5
d=1/2(5.5)(5.5)^2=83m
w f=2200(5.5)=12100 x 83=1004300
p= 1004300/5.5
Answer: 182600 W
answer is 7
Keep in mind work can equal ∆KE too.
w = ∆KE = final KE - initial KE =
(0.5 * 2200 * 30^2) - 0 = 990000 J
p = w/t = 990000/5.5 = 180000 W
I think I'm in the same class mssailormouth had in 2019,
and I believe this is how the problem was intended to be
solved.
w = ∆KE = final KE - initial KE =
(0.5 * 2200 * 30^2) - 0 = 990000 J
p = w/t = 990000/5.5 = 180000 W
I think I'm in the same class mssailormouth had in 2019,
and I believe this is how the problem was intended to be
solved.
1 answer