Question

An object initially at rest moves 6034 m in 3 hours. What acceleration occurred for the object to achieve that distance?

A) 0.01 m/s²
B) 4 m/s²
C) 6 m/s²
D) 12 m/s²

Answer 1

The acceleration that occurred for an object to move 6034 m in 3 hours from rest is approximately 0.01 m/s², which corresponds to option A) 0.01 m/s².

Step-by-step explanation:

To determine the acceleration that occurred for an object to move 6034 m in 3 hours (10800 seconds), we can use the kinematic equation that relates distance traveled to initial velocity, time, and acceleration. The equation is s = ut + ½at², where s is the distance, u is the initial velocity, t is the time, and a is the acceleration. Given that the initial velocity (u) is zero since the object starts from rest, we can simplify the equation to s = ½at².

Plugging in the given values, we have 6034 m = ½ * a * (10800 s)². Solving for a gives us a = 2 * 6034 m / (10800 s)², which calculates to approximately 0.01 m/s².

Therefore, the correct answer is A) 0.01 m/s².