Question

two objects, x and y, accelerate from rest with the same constant acceleration. object x accelerates for twice the time as object y. which of the following is true of these objects at the end of their respective periods of acceleration?

Answer 1

Answer:

X has travelled 4 times as far as Y

X is moving 2 times as fast as Y

Explanation:

Because x=1/2at²(get rid of the first portion of x=vt+1/2at² because initial v is 0 which multiplies to 0). If the acceleration is set to for example 10 (keep this constant for both X and Y) and we use t=1 and t=2 for time (as the time travelled of X is twice that of Y). Plugging this into the equation, at t=1, x=4. At t=2, x=20. 20 is four times greater than 5, so X has travelled 4 times as far as Y.

To find the difference in speed between the two objects, use the equation $v_{f}=v_{0} +at$. Since the initial velocity is 0, that part can just be removed from the equation. With v=at, it is easy to see that if the time plugged in is twice for one than the other (and the acceleration is the same for both), the final result will be twice of the other as well. For example: If the acceleration is 10 again for both, then v=10t. If t is 1, the velocity is 10. If t is 2, the velocity is 20.