Question

A spaceship moving with an initial velocity of 58.0 meters/second experiences a uniform acceleration and attains a final velocity of 153 meters/second. What distance has the spaceship covered after 12.0 seconds?

Answer 1

To determine the distance covered by the spaceship after 12 seconds, we calculate the acceleration first, which is 7.92 m/s², and then use it to find the distance, which is 1,391.52 meters.

Step-by-step explanation:

The distance covered by a spaceship that moves with an initial velocity of 58.0 meters/second, experiences a uniform acceleration, and attains a final velocity of 153 meters/second after 12.0 seconds can be calculated using the equation of motion:

S = ut + ½at²

where:

• S is the distance covered,
• u is the initial velocity,
• a is the acceleration,
• t is the time elapsed.

First, we find the acceleration (a) using the equation:

v = u + at

We rearrange this to solve for a:

a = (v - u) / t

By substituting the known values:

a = (153 m/s - 58 m/s) / 12 s = 7.92 m/s²

Now that we have the acceleration, we can calculate S:

S = (58 m/s × 12 s) + ½(7.92 m/s² × (12 s)²) = 1,391.52 m

The spaceship has covered a distance of 1,391.52 meters after 12 seconds.

Answer 2

First, we determine the acceleration by using the given initial and final velocities and the time.
acceleration = (153 m/s - 58 m/s) / 12 s
= 7.917 m/s²
Then, use the equation below to solve for the distance.
distance = (V² - v²) / 2a
where V is the final velocity, v is the initial velocity and a is the acceleration. Substituting,
d = (153² - 58²) / 2(7.917)
= 1266 m
Thus, the spaceship have traveled 1,266 m.