A boy sliding down a hill accelerates at 1.40m/s^2... physics question.?

N

Noura

New member
first you need to use a kinematic equation
Vf^2=Vo^2+2a(change x)
you are looking for change in x
Vf=9
Vo=0
A=1.4
once you look at the problem and know what you have you can plug in the numbers
9^2=0^2+2(1.4)(change in x)
81=0+2.8(change in x)
change in x= 28.93 29(rounded)
distance =28.93 m or 29m (rounded)

i hope this helps
 
first you need to use a kinematic equation
Vf^2=Vo^2+2a(change x)
you are looking for change in x
Vf=9
Vo=0
A=1.4
once you look at the problem and know what you have you can plug in the numbers
9^2=0^2+2(1.4)(change in x)
81=0+2.8(change in x)
change in x= 28.93 29(rounded)
distance =28.93 m or 29m (rounded)

i hope this helps
 
A boy sliding down a hill accelerates at 1.40 m/s^2. If he started from rest, in what distance would he reach a speed of 9.00 m/s?


Thanks in advance!
 
first you need to use a kinematic equation
Vf^2=Vo^2+2a(change x)
you are looking for change in x
Vf=9
Vo=0
A=1.4
once you look at the problem and know what you have you can plug in the numbers
9^2=0^2+2(1.4)(change in x)
81=0+2.8(change in x)
change in x= 28.93 29(rounded)
distance =28.93 m or 29m (rounded)

i hope this helps
 
first you need to use a kinematic equation
Vf^2=Vo^2+2a(change x)
you are looking for change in x
Vf=9
Vo=0
A=1.4
once you look at the problem and know what you have you can plug in the numbers
9^2=0^2+2(1.4)(change in x)
81=0+2.8(change in x)
change in x= 28.93 29(rounded)
distance =28.93 m or 29m (rounded)

i hope this helps
 
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