Question

A rocket ship undergoes an acceleration of 112 m/s2 for a period of 8 seconds. Knowing the rocket began with an initial velocity of 1000 m/s, how far did it go during the 8 second period of acceleration?

Options:
A) 800 m
B) 896 m
C) 1008 m
D) 1120 m

Answer 1

During the 8-second period of acceleration, given an initial velocity of 1000 m/s and an acceleration of 112 m/s^2, the rocket ship traveled a distance of 11584 m. This is calculated using the kinematic equation s = ut + ½at^2.

Step-by-step explanation:

The question provided asks us to determine the distance a rocket ship has traveled during an 8-second period of acceleration, given that it starts with an initial velocity of 1000 m/s and accelerates at a rate of 112 m/s2. To find the distance, we can use the kinematic equation for uniformly accelerated motion:

s = ut + ½at2

Where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.

Plugging in the given values:

s = (1000 m/s)(8 s) + ½(112 m/s2)(8 s)2
s = 8000 m + ½(112)(64) m
s = 8000 m + 3584 m
s = 11584 m

Therefore, the rocket ship traveled a distance of 11584 m during the 8-second period of acceleration. Since this value is not one of the options provided, it's possible that there may have been a typo or mistake in the set of options given for the answer.