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A top fuel drag racer reaches a maximum speed of 330 mph at the end of the 1/4-mile (402 m) racetrack. Part A - Assuming that the acceleration is constant during the race.

a. Calculate the average value of the acceleration of the top fuel car.
b. What is the ratio of that acceleration to the acceleration due to gravity?

Respuesta :

adelekeademolu

Answer:

a. 27 m/s2

b. 2.75:1

Explanation:

a.

Since it is assumed that the race car starts from rest, initial velocity, u = 0 m/s

At the end of 402 m, distance, s = 402 m; and

Final velocity, v = 330 m/h

Converting this to m/s,

[tex]v=\dfrac{330\times1609}{3600}=147.5[/tex] m/s

Using the equation of motion

[tex]v^2=u^2+2as[/tex]

[tex]a=\dfrac{v^2-u^2}{2s}[/tex]

[tex]a=\dfrac{147.5^2-0^2}{2\times402}=27[/tex]

b.

If acceleration due to gravity is taken as 9.8 m/s2, then

[tex]a:g=27:9.8=2.75:1[/tex]

Eduard22sly

A. The average value of the acceleration of the top fuel car is 27.06 m/s²

B. The ratio of the acceleration of the car to the acceleration due to gravity is 2.76 : 1

A. How to determine the acceleration of the car

From the question given above, the following data were obtained:

  • Initial velocity (u) = 0 m/s
  • Final velocity (v) = 330 mph = 330 × 0.447 = 147.51 m/s
  • Distance (s) = 402 m
  • Acceleration (a) =?

v² = u² + 2as

147.51² = 0² + 2 × a × 402

21759.2001 = 804a

Divide both side by 804

a = 21759.2001 / 804

a = 27.06 m/s²

B. How to determine the ratio of the acceleration of the car to the acceleration due to gravity

  • Acceleration of car (a) = 27.06 m/s²
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Ratio (a/g) =?

Ratio = a/ g

Ratio = 27.06 / 9.8

Ratio = 2.76 : 1

Learn more about acceleration:

https://brainly.com/question/3044652

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