High School

A cheetah reaches a final velocity of 39.4 ft/sec in 1.25 seconds. If it accelerated at a rate of 2.10 ft/sec\(^2\), what was its initial velocity, and how far did it travel in this time?

Answer :

Final answer:

The cheetah's initial velocity was 36.9 ft/sec, and it traveled approximately 61.5 ft in 1.25 seconds.

Explanation:

To find the cheetah's initial velocity, we can use the formula:

v = u + at

Where:

v is the final velocity (39.4 ft/sec)u is the initial velocity (unknown)a is the acceleration (2.10 ft/sec²)t is the time taken (1.25 seconds)

Substituting the given values into the formula, we get:

39.4 = u + 2.10 × 1.25

Solving for u, we find:

u = 36.9 ft/sec

To find the distance traveled by the cheetah, we can use the formula:

s = ut + 0.5at²

Substituting the given values into the formula, we get:

s = 36.9 × 1.25 + 0.5 × 2.10 × (1.25)²

s ≈ 61.5 ft

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