Solving Acceleration Problem for Jaguar S-Type R

In summary, the speaker is expressing gratitude for previous help in physics and brings up a new question regarding a project involving a car's power and kinetic energy. They provide the necessary variables and question if they can calculate the force by converting power into Newtons. However, it is suggested to use the work-energy theorem to calculate the time it takes to reach a certain power instead.
  • #1
tucky
30
0
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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  • #2
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
 
  • #3
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".
 

1. How do you calculate acceleration for a Jaguar S-Type R?

To calculate acceleration for a Jaguar S-Type R, you will need to use the formula: acceleration = (final velocity - initial velocity) / time. The final velocity can be determined by using a GPS or a speedometer, and the initial velocity can be assumed to be 0 mph if the car is starting from a standstill. Time can be measured using a stopwatch or a timer.

2. What is the average acceleration for a Jaguar S-Type R?

The average acceleration for a Jaguar S-Type R can vary depending on factors such as road conditions, driver behavior, and engine performance. However, on average, the Jaguar S-Type R has an acceleration of 0-60 mph in 5.3 seconds.

3. How can I improve the acceleration of my Jaguar S-Type R?

To improve the acceleration of your Jaguar S-Type R, you can consider upgrading the engine or adding performance parts such as a cold air intake or a performance exhaust system. Additionally, regular maintenance and tune-ups can also help improve the acceleration of your car.

4. What are common acceleration problems in Jaguar S-Type R?

Some common acceleration problems in Jaguar S-Type R include engine misfires, issues with the fuel system, and transmission problems. It is important to have these issues diagnosed and fixed by a certified mechanic to ensure optimal performance of your car.

5. How does acceleration affect the overall performance of a Jaguar S-Type R?

Acceleration plays a crucial role in the overall performance of a Jaguar S-Type R. A high acceleration rate can result in faster speeds and better handling, while a slow acceleration rate can affect the overall driving experience. Additionally, acceleration can also impact the fuel efficiency and engine wear of the car.

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