Solving Acceleration Problem for Jaguar S-Type R

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The discussion revolves around calculating the time required for a Jaguar S-Type R to accelerate from 0 to various speeds using physics principles. The user has determined the car's power output, mass, and kinetic energy, and seeks to calculate the force and acceleration. By applying the work-energy theorem, they find the work done to reach 30 mph and subsequently calculate the time needed to achieve that speed. The key takeaway is that the user effectively used conservation of energy rather than directly calculating force. This approach led to the conclusion that it takes approximately 1.3 seconds to accelerate from 0 to 30 mph.
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First off, thanks to all who helped me on my last questions. It is greatly appreciated that you are willing to help alleviate my physics stress! I am sure there will be many more questions to come. In fact, here is another physic's predicament that I am stranded over.

Background: I am doing a project dealing with a car (Jaguar S-Type R). I have calculated the power (P = E/t) and the Kinetic energy (KE = 1/2mv2)

Question: If the engine generates a constant power of Pavg, calculate the time it would take to accelerate from 0-30mph, 0-40mph, etc.

I know the speeds (v0 = 0mph, v = 30mph), mass (1847.48kg), power (171.3078Hp, or 94219.284 ft-lbs/s), and Kinetic energy (664,537.1725J)

So, can I calculate the Force by converting 94219.284ft-lbs/s into Newtons? I really do not know what ft-lbs/s can be converted. If I was able to determine the force of this car than I could use Newton’s second law equation: acceleration (Sum of Force)/(time) this would determine the acceleration; and then I could determine the sec by using this equation: a = change in v/ time

I think I am completely off track, if anyone could help me it would be greatly appreciated!
 
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Power = work / time = W/t

You state the power is equal to 171.3078Hp, which is equal to 127.744kW

v = 30mph = 13.4112m/s

K = (1/2)*m*v^2 = .5 * 1847.48 * 13.4112^2 = 166,144J

Using the work-energy theorem W = K2 - K1, where K2 = 166,144J and K1 = 0J.

W = 166,144J

Therefore the time it takes to reach the power (or go from 0 to 30mph) is:

t = W/P = 166,144J / 127.744kW = 1.3s
 
Notice that dduardo did NOT calculate force!

Since power (i.e. "horsepower") is work/time (energy/time), he
calculated the change in kinetic energy between "0-30mph, 0-40mph, etc." and then calculated how much time was required to give that much energy. In other words, he used "conservation of energy".
 
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