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Jeff Gordon accelerates his race car at 2.7 m/s² and covers a distance of 400 m. What is his final velocity at the end of the 400 m distance?

a) 10.8 m/s
b) 400.0 m/s
c) 1080 m/s
d) 648.0 m/s

1 Answer

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Final answer:

The final velocity of Jeff Gordon's race car at the end of the 400 m distance is approximately 46.48 m/s, calculated using the kinematic equation v^2 = u^2 + 2as, where u=0, a=2.7 m/s², and s=400 m.

Step-by-step explanation:

To find the final velocity of Jeff Gordon's race car, we need to use the kinematic equation which relates acceleration, initial velocity, distance, and final velocity:

v2 = u2 + 2as

Where v is the final velocity, u is the initial velocity (which is 0 since he starts from rest), a is the acceleration, and s is the distance covered.

Since the initial velocity (u) is 0, the equation simplifies to v2 = 2as. Plugging in the values:

v2 = 2 * 2.7 m/s2 * 400 m

v2 = 2160 m2/s2

v = √2160 m2/s2

v ≈ 46.48 m/s

Therefore, the final velocity at the end of the 400 m distance is approximately 46.48 m/s. This is not one of the provided options, implying there might be a typo in the answer choices.

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