Ask Question
24 June, 00:45

Car drag racing takes place over a distance of a 1 4 mile (402 m) from a standing start. If a car (mass 1500 kg) would be propelled forward with a pulling force equal to that of gravity, what would be the change in kinetic energy and the terminal speed of the car (in mph) at the end of the race be?

+3
Answers (1)
  1. 24 June, 01:15
    0
    The change in kinetic energy is 5,909,400J and the terminal speed of the car is 160,800mph

    Explanation:

    Change in kinetic energy = mg (h2 - h1)

    Mass (m) = 1500kg, g = 9.8m/s^2, final distance (h2) = 402m, initial distance (h1) = 0m

    Change in kinetic energy = 1500*9.8 * (402 - 0) = 1500*9.8*402 = 5,909,400J

    From equations of motion

    h = ut + 1/2gt^2 [u is initial speed and is equal to zero because the car drag racing began from a standing start (rest) ]

    h = 1/2gt^2

    402 = 1/2 * 9.8t^2

    402 = 4.9t^2

    t^2 = 402/4.9 = 82.04

    t = √82.04 = 9.06s * 1h/3600s = 0.0025h

    Terminal speed = distance/time = 402m/0.0025h = 160,800mph
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Car drag racing takes place over a distance of a 1 4 mile (402 m) from a standing start. If a car (mass 1500 kg) would be propelled forward ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers