1.)A car travelling at 15m/s , slows down uniformly at rest in 6.0
a)how far does the car travel during the 6.0 s
seconds?
b)what is the car travel during the 6.0 seconds?
2.)A car takes 10 s to go from 0m/s to 24 m/s at constant acceleration
a)what is the value of a
b)find the distance for this
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OpenStudy (anonymous):
for the 1st one i can do distance like what we used earlier right
OpenStudy (isaiah.feynman):
We need acceleration first. Its not 9.8 this time.
OpenStudy (anonymous):
so since velocity is change i velocity/ time we use this eq
OpenStudy (isaiah.feynman):
Okay we will use this equation. \[D=\frac{ 1 }{ 2 }(u+v)t \]Now can you show me how to use that equation?
OpenStudy (anonymous):
yes:)
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OpenStudy (isaiah.feynman):
Okay what is the initial velocity?
OpenStudy (anonymous):
ok would it be 1/2(1+6)
OpenStudy (anonymous):
initial v is 15
OpenStudy (isaiah.feynman):
Yes, and final velocity?
OpenStudy (anonymous):
6.0
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OpenStudy (isaiah.feynman):
6.0 is time (In the question). Final velocity is 0 "slows down uniformly at rest"
OpenStudy (anonymous):
kk initial velocity is 15 and the final is 0 so 1/2 (15+0)6.0
OpenStudy (isaiah.feynman):
Yes.
OpenStudy (anonymous):
k i got 45
OpenStudy (isaiah.feynman):
Correct!
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OpenStudy (anonymous):
yay:)
OpenStudy (anonymous):
so for be all i need to do is d/t right which is 15/6.0?
OpenStudy (isaiah.feynman):
Brb gotta do stuff in the house.
OpenStudy (anonymous):
kk
OpenStudy (isaiah.feynman):
I don't quite understand the 1b
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OpenStudy (anonymous):
oh its suppose to say what is the cars average velocity during 6.0 seconds .i got 2.5
OpenStudy (isaiah.feynman):
The average velocity is the distance/time= 45/6=7.5m/s
OpenStudy (anonymous):
smt yeah i got that t dnt know why i typed that
OpenStudy (isaiah.feynman):
Its okay.
OpenStudy (anonymous):
for 2 a its acceeration so i can use vf-vi/t?
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OpenStudy (isaiah.feynman):
That's what you use.
OpenStudy (anonymous):
im right??
OpenStudy (isaiah.feynman):
Yes you are.
OpenStudy (anonymous):
yay i got 25-0/10=2.5
OpenStudy (isaiah.feynman):
24 not 25 so a=2.4
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OpenStudy (anonymous):
how come i get 2.5 in my calculator?
OpenStudy (isaiah.feynman):
Just a typo.
OpenStudy (anonymous):
kk so if i fint distance it would be 1/2 at^2
OpenStudy (isaiah.feynman):
Yes. Note that a=2.4
OpenStudy (anonymous):
y do i keep geting 2.5 isaiah???:(
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OpenStudy (isaiah.feynman):
The question says "A car takes 10 s to go from 0m/s to 24 m/s at constant acceleration"
OpenStudy (anonymous):
1/2 (5)(0)(25)
OpenStudy (isaiah.feynman):
1/2(2.5)(100)
OpenStudy (anonymous):
y u did use 10 s instead of 100?
OpenStudy (isaiah.feynman):
t^2. t is 10, t^2 is 10^2=100
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OpenStudy (anonymous):
ohhhhhhhh k i get it
OpenStudy (anonymous):
k thanks alot u r very patient
OpenStudy (isaiah.feynman):
Thank you.
OpenStudy (isaiah.feynman):
What's your name?
OpenStudy (anonymous):
rissa
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