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Drupady [299]
4 years ago
7

How long will it take a cyclist with a forward acceleration of -0.50 m/s squared to bring a bicycle with an initial velocity of

forward of 13.5 m/s to a complete stop? show the formula and your work.
Physics
1 answer:
Liono4ka [1.6K]4 years ago
3 0
Using kinematic equation v=u+at......... 1.
v = final velocity = 0 m/s. u = initial velocity = 13.5 m/s.
a = -0.5 m/s². 
Plug all the known values in 1 to find t
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3. In 1989, Michel Menin of France walked on a tightrope suspended under a
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Answer: 80m

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Using the equation of motion,

V^2= U^s + 2gs--------2

U= initial speed is 0m/s

g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s

40m/s is contant since U (the coin is at rest is 0) hence V =40m/s

Slotting our values into equation 2

40^2= 0^2 + 2 * 10* (3150-y)

1600 = 0 + 63000 - 20y

1600 - 63000 = - 20y

-61400 = - 20y minus cancel out minus on both sides of the equation

61400 = 20y

Hence y = 61400/20

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x = 3150 - 3070

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