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igor_vitrenko [27]
3 years ago
15

A hockey player hits a puck so that it comes to rest in 9 s after sliding 30 m on the ice. Determine (a) the acceleration in ter

ms of ,​(4 marks) (b) the initial velocity and acceleration of the puck.​(5 marks)
Physics
1 answer:
blondinia [14]3 years ago
4 0

Answer:

a) Acceleration, a = -u/9, where u is the initial velocity

b) Acceleration = 0.741 m/s², Initial velocity = 6.67 m/s

Explanation:

 We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

a)  We have v = 0 m/s, t = 9 s

 So,    0 = u + a x 9

      Acceleration, a = -u/9, where u is the initial velocity

b) We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 s = 30 m , u = -9a, t =9 s

 So,  30= -9a*9+\frac{1}{2} *a*9^2\\ \\ 30=-40.5a\\ \\ a=-0.741m/s^2

u = -9a = 6.67 m/s

So acceleration = 0.741 m/s², Initial velocity = 6.67 m/s

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x=\frac{x^{|}+vt^{|} }{\sqrt{1-\frac{v^{2} }{c{2} } } } \\x=\frac{2m+0.92c(0) }{\sqrt{1-\frac{(0.92c)^{2} }{c{2} } } }\\x=5.103m\\y=y^{|}\\ y=3.7m\\z=z^{|}\\ z=3.7m

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r=\frac{1}{\sqrt{1-v^{2} } } \\r=\frac{1}{\sqrt{1-(0.92)^{2} } } \\r=2.551\\t=r(t^{|}+vx^{|}/c^{2}   )\\t=2.551(0s+(0.92c)(2)/c^{2} )\\t=1.57*10^{-8}s

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