What Is the Average Power Output of a Race Car's Motor During Acceleration?

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The discussion focuses on calculating the average power output of a 1250-kg race car accelerating from rest to 30.0 m/s in 4.00 seconds on a frictionless surface. The relevant formulas include power (P = E/t), force (F = ma), and kinetic energy (Ek = 1/2 mv^2). Participants clarify that the change in energy should be kinetic, and a mistake was made by incorrectly using 15 m/s instead of the correct 30 m/s in calculations. The correct average power output is determined to be 141 kW. The conversation emphasizes the importance of accurate values in physics calculations.
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A 1250-kg race car accelerates uniformly from rest to 30.0m/s in 4.00s on a horizontal surface with no friction. What must be the average power output of its motor? (Ans:141kW)

--
P=E/t
F=ma
Ek=1/2mv^2

--

P=E/t
P=Ek/t
=(.5X1250kgX15.0m/s^2) / (4.00s)
=35156.25 W

How exactly do I start this off? I assume that the change in energy should be kinetic only, right?.
 
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k0k said:
A 1250-kg race car accelerates uniformly from rest to 30.0m/s in 4.00s on a horizontal surface with no friction. What must be the average power output of its motor? (Ans:141kW)

--
P=E/t
F=ma
Ek=1/2mv^2

--

P=E/t
P=Ek/t
=(.5X1250kgX15.0m/s^2) / (4.00s)
=35156.25 W

How exactly do I start this off? I assume that the change in energy should be kinetic only, right?.

I think you would have had the correct answer had you not replaced the given 30 m/s with 15 m/s.
 
OH, okay got it now. : D
Thanks
 

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