Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,
Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?
a = -1.47 m/ (a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/ (since same friction force is applied)
s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
I have no idea I need the answer too
Either one is fun and great to play!
<u>Answer</u>
(g²n - m)/(gm)
<u>Explanation</u>
g - m ÷ gn = g - m/gn
Make the equation have the same denominator
g - m ÷ gn = g - m/gn = (ggn)/gn - m/gn
= (g²n)/gn - m/gm
Since they have the same denominator, we can carry out the subtraction on the numerator and then put them under one denominator.
(g²n)/gn - m/gm = (g²n - m)/(gm)
