Final answer:
The object with twice its initial velocity will travel four times the distance, and the object with four times its initial velocity will travel sixteen times the distance, when both have constant acceleration.
Step-by-step explanation:
To compare the distance traveled by two objects with different changes in velocity over the same time period with constant acceleration, one must understand the kinematic equation v^2 = u^2 + 2a(x - x_0), where v is the final velocity, u is the initial velocity, a is the acceleration, and (x - x_0) is the displacement. The final velocity v, in each case, will be proportional to their respective changes in velocity as the initial velocities are multiplied by a factor (twice for one object and four times for another object).
Using the kinematic equation, if you solve for displacement while doubling or quadrupling the initial velocity, the displacement (or distance traveled) will in
crease by a factor of four or sixteen, respectively. Therefore, if v is twice u, then the object will travel four times the distance, while if v is four times u, it will travel sixteen times the distance assuming the time is constant.
In summary, for these two objects under constant acceleration:
- The object with twice its initial velocity will travel four times the distance.
- The object with four times its initial velocity will travel sixteen times the distance.
Therefore, the correct answer is c) The object with four times the initial velocity will travel sixteen times the distance.