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A rocket initially at rest accelerates at a rate of 99.0 meters/second2. Calculate the distance covered by the rocket if it attains a final velocity of 445 meters/second after 4.50 seconds.

A2.50 × 102 meters
B1.00 × 103 meters
C5.05 × 102 meters
D2.00 × 103 meters
E1.00 × 102 meters

Verified Answer :

Answer: The correct answer is option b.

Explanation: We are given that the rocket is at rest initially final velocity is 445m/s.

The acceleration of the rocket is \(99.0m/s^2\)

To calculate the distance of rocket, we use third equation of motion, which is:

\(v^2-u^2=2as\)

where, v = final velocity = 445m/s

u = initial velocity = 0m/s

a = acceleration = \(99m/s^2\)

s = distance = ? m

Putting values in above equation, we get:

\((445)^2-(0)^2=2(99)s\\\\s=1\times 10^3meters\)

ghostcd40

Answer:

the answere is B

Explanation:

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