Respuesta :
The average power needed to produce this final speed is 1069.1 hp.
Mass of the dragster, Â m = 500.0 kg,
Displacement travelled by the dragster, Â s = 402 m,
Time taken in this travel, Â t = 5.0 s,
Final velocity of the dragster, Â v = 130 m/s.
Let the initial velocity of the dragster be u and acceleration be a.
Using kinematical equation, Â s = ut + (1/2)at^2.
402 Â = Â u*5 Â + (1/2)*a*5^2
10*u + 25*a  = 804.    ...........(1)
Using kinematical equation, v = u +at.
130 = u + 5*a
5*u + 25*a = 650. Â Â Â .............(2)
Solving (1)and (2), we get,
u = Â 30.8 m/s.
According to work-energy theorem,
Work done = change in kinetic energy
W Â = (1/2)*m*(v^2 - u^2)
W = (1/2)*500*(130^2 - 30.8^2)
W Â = Â 3987840. J
Therefore power rating of the dragster is given by,
P  ⇒  W/t. =  3987840/5 = 797568 watt.
P  ⇒ 797568/746 =  1069.1 hp.
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The answer is 1069.1 hp. Â
Mass of the dragster, Â m = 500.0 kg,
Displacement travelled by the dragster, Â s = 402 m,
Time taken in this travel, Â t = 5.0 s,
Final velocity of the dragster, Â v = 130 m/s.
Let the initial velocity of the dragster be u and acceleration be a.
Using kinematical equation, Â s = ut + (1/2)at^2.
 402  =  u*5  + (1/2)*a*5^2
 10*u + 25*a  = 804.    ...........(1)
Using kinematical equation, v = u +at^.
 130 = u + 5*a^
5*u + 25*a = 650. Â Â Â .............(2)
Solving (1)and (2), we get,
u = Â 30.8 m/s.
According to work -energy theorem,
Work done = change in kinetic energy
 W  = (1/2)*m*(v^2 - u^2)
, Â W = (1/2)*500*(130^2 - 30.8^2)
, Â W Â = Â 3987840. J
therefore Power rating of the dragster is given by,
P Â = Â W/t. = Â 3987840/5 = 797568 watt.
, Â P Â = 797568/746 = Â 1069.1 hp. Â Â Â Â
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