The average power needed to produce this final speed is 1069.1 hp.
Mass of the dragster, m = 500.0 kg,
Displacement travelled by the dragster, s = 402 m,
Time taken in this travel, t = 5.0 s,
Final velocity of the dragster, v = 130 m/s.
Let the initial velocity of the dragster be u and acceleration be a.
Using kinematical equation, 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. .............(2)
Solving (1)and (2), we get,
u = 30.8 m/s.
According to work-energy theorem,
Work done = change in kinetic energy
W = (1/2)*m*(v^2 - u^2)
W = (1/2)*500*(130^2 - 30.8^2)
W = 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.
Mass of the dragster, m = 500.0 kg,
Displacement travelled by the dragster, s = 402 m,
Time taken in this travel, t = 5.0 s,
Final velocity of the dragster, v = 130 m/s.
Let the initial velocity of the dragster be u and acceleration be a.
Using kinematical equation, 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. .............(2)
Solving (1)and (2), we get,
u = 30.8 m/s.
According to work -energy theorem,
Work done = change in kinetic energy
W = (1/2)*m*(v^2 - u^2)
, W = (1/2)*500*(130^2 - 30.8^2)
, W = 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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