51.3k views
5 votes
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. A. 2.50 x 10² meter B. 1.00 x 10³ meters C. 5.05 x 10² meters D. 2.00 x 10³ meters E. 1.00 x 10² meters

User Damned
by
7.2k points

1 Answer

5 votes

Final answer:

To find the distance covered by a rocket accelerating at 99.0 meters/second² to 445 meters/second over 4.50 seconds, we used the kinematic equation s = v_0 ⋅ t + ½ at², which yielded a result of approximately 1001.25 meters or 1.00 x 10³ meters in scientific notation. The correct answer is option B. 1.00 x 10³ meters.

Step-by-step explanation:

The student's question asks us to calculate the distance covered by a rocket accelerating at a rate of 99.0 meters/second² until it reaches a final velocity of 445 meters/second after accelerating for 4.50 seconds.

To solve this problem, we can use the kinematic equation for uniformly accelerated motion:

s = v_0 ⋅ t + ½ at²

Where:

s is the distance covered

v_0 is the initial velocity (which is 0 since the rocket starts at rest)

a is the acceleration

t is the time

Plugging the values into the equation, we get:

s = 0 ⋅ 4.50 s + ½ ⋅ 99.0 m/s² ⋅ (4.50 s)²

Calculating the covered distance:

s = 0 + ½ ⋅ 99.0 m/s² ⋅ 20.25 s²

s = ½ ⋅ 99.0 m/s² ⋅ 20.25 s²

s = 49.5 ⋅ 20.25 m

s = 1001.25 meters

Therefore, the nearest option to the distance covered by the rocket expressed in scientific notation is B. 1.00 x 10³ meters.

User Nicholas Piasecki
by
7.3k points