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

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SUMMARY

The average power output of a 1250-kg race car's motor during acceleration from rest to 30.0 m/s in 4.00 seconds is calculated to be 141 kW. The relevant formulas used include power (P = E/t), force (F = ma), and kinetic energy (Ek = 1/2 mv^2). The calculation involves determining the change in kinetic energy over time, confirming that the correct final velocity is 30 m/s, not 15 m/s. This discussion highlights the importance of accurate values in physics calculations.

PREREQUISITES
  • Understanding of Newton's Second Law (F = ma)
  • Knowledge of kinetic energy formula (Ek = 1/2 mv^2)
  • Familiarity with power calculation (P = E/t)
  • Basic principles of uniform acceleration
NEXT STEPS
  • Study the derivation of kinetic energy and its applications in real-world scenarios.
  • Learn about the implications of power output in automotive engineering.
  • Explore advanced topics in dynamics, such as frictionless motion and its effects on acceleration.
  • Investigate the relationship between power, speed, and acceleration in different vehicle types.
USEFUL FOR

Physics students, automotive engineers, and anyone interested in the mechanics of race car performance will benefit from this discussion.

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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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