Final answer:
This high school physics question is about analyzing an object's acceleration and velocity over time using kinematic equations. Given the time-varying acceleration a = (2 m/s²)t, we can use a = Δv / Δt to calculate acceleration with known changes in velocity and time, and understand motion through velocity-time and acceleration-time graphs.
Step-by-step explanation:
The question involves analyzing an object initially at rest that experiences a time-varying acceleration given by a = (2 m/s²)t. We're looking at a problem where the object stays at rest until a certain time, then begins to accelerate in a positive direction and eventually moves with a constant velocity.
To solve for acceleration when given initial and final velocities and elapsed time, we use the kinematic equation a = Δv / Δt, where Δv is the change in velocity and Δt is the change in time.
For example, if an object starts from rest and reaches a final velocity of 8.00 m/s in 2.50 s, we would find the acceleration by taking Δv = 8.00 m/s and Δt = 2.50 s, giving us an acceleration of a = 8.00 m/s / 2.50 s = 3.20 m/s².
To address a practical example from kinematic equations: If a body starts from rest and accelerates at 4 m/s² for 2 s, the final velocity is indeed 8 m/s, because you simply multiply the acceleration by the time (v = at).
As for the acceleration vs. time graph, since the acceleration varies with time, the graph would show a slope increasing over time, and the area under the graph up to a certain time would give the final velocity at that particular time. When the object travels with a constant velocity, the acceleration is zero, and thus the acceleration-time graph would show a horizontal line during that period.