Answered

Select the correct answer.
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.
O A.
2.50 × 102 meters
O B.
1.00 × 10³ meters
○ C. 5.05 x 102 meters
O D. 2.00 × 103 meters
O E.
1.00 x 10² meters

Verified Answer :

GinnyAnswer

To solve this problem, we need to use the kinematic equations of motion. Specifically, we can use the following two kinematic equations to find the distance covered by the rocket:

1. \(\( v = u + at \)\)
2. \(\( s = ut + \frac{1}{2}at^2 \)\)

Where:
- \(\( v \)\) = final velocity (445 m/s)
- \(\( u \)\) = initial velocity (0 m/s, since the rocket is initially at rest)
- \(\( t \)\) = time (4.50 seconds)
- \(\( a \)\) = acceleration (99.0 m/s²)
- \(\( s \)\) = distance

First, confirm the final velocity using the first equation:

\(\[ v = u + at \]\)

Given:
\(\[ u = 0 \, \text{m/s} \]\)
\(\[ a = 99.0 \, \text{m/s}^2 \]\)
\(\[ t = 4.50 \, \text{s} \]\)

\(\[ v = 0 + (99.0 \times 4.50) \]\)
\(\[ v = 445.5 \, \text{m/s} \]\)

This matches approximately with the given final velocity (445 m/s), allowing us to use these values confidently.

Now, use the second equation to calculate the distance \(\( s \)\):

\(\[ s = ut + \frac{1}{2}at^2 \]\)

Given:
\(\[ u = 0 \, \text{m/s} \]\)
\(\[ a = 99.0 \, \text{m/s}^2 \]\)
\(\[ t = 4.50 \, \text{s} \]\)

\(\[ s = (0 \times 4.50) + \frac{1}{2} (99.0) (4.50)^2 \]\)
\(\[ s = 0 + 0.5 \times 99.0 \times 20.25 \]\)
\(\[ s = 49.5 \times 20.25 \]\)
\(\[ s = 1001.25 \, \text{meters} \]\)

Hence, the distance covered by the rocket is approximately 1000 meters.

Therefore, the correct answer is:

B. 1.00 × 10³ meters

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