Question

During a 29.4-s interval, the velocity of a rocket changes from 204 m/s to 508m/s. Assuming constant acceleration, determine the displacement of the rocket during this time interval.

Answer 1

The displacement of the rocket during the 29.4-second interval, given an initial velocity of 204 m/s and a final velocity of 508 m/s, is 10460.8 meters.

Step-by-step explanation:

To determine the displacement of a rocket during a 29.4-second interval with an initial velocity of 204 m/s and a final velocity of 508 m/s assuming constant acceleration, we can use the kinematic equations for uniformly accelerated motion. The equation that relates initial velocity (vo), final velocity (v), acceleration (a), and displacement (s) is:

s = vo t + ½ a t2

First, we need to calculate the acceleration using the formula:

a = (v - vo) / t

a = (508 m/s - 204 m/s) / 29.4 s = 10.34 m/s2

Now, we substitute the known values into the displacement formula:

s = (204 m/s)(29.4 s) + ½(10.34 m/s2)(29.4 s)2

s = 5997.6 m + 0.5 (10.34 m/s2)(864.36 s2)

s = 5997.6 m + 4463.2 m

s = 10460.8 m

The displacement of the rocket during the 29.4-second interval is 10460.8 meters.