Question

suppose you are looking at two graphs of position vs time. the first graph is of an object undergoing constant positive velocity of 2 m/s, and the second graph is of an object undergoing constant positive acceleration of 2 m/s². how do the graphs compare? (1 point) both graphs increase with time, but the first graph curves upward, becoming steeper and steeper, while the second graph is linear with a positive slope. both graphs should be linear with the same positive slope. both graphs increase with time, but the first graph is linear with a positive slope while the second graph curves upward, becoming steeper and steeper. the first graph is a constant horizontal line while the second is linear and slopes upward.

Explanation:

Step1: Recall velocity - position relationship

Velocity $v=\frac{\Delta x}{\Delta t}$. For constant velocity $v = 2\ m/s$, the position - time graph has a constant slope. The position $x$ as a function of time $t$ is given by $x=x_0+vt$ (where $x_0$ is the initial position), which is a linear equation of the form $y = mx + c$ with $m = v$ (positive slope).

Step2: Recall acceleration - position relationship

Acceleration $a=\frac{\Delta v}{\Delta t}$. For constant acceleration $a = 2\ m/s^2$, the velocity $v=v_0 + at$ and the position $x=x_0+v_0t+\frac{1}{2}at^2$. Since the position is a quadratic function of time ($x$ is proportional to $t^2$), the graph of $x$ vs $t$ is a parabola opening upwards. As time increases, the slope of the tangent to the curve (which represents velocity) becomes steeper and steeper.

Answer:

C. Both graphs increase with time, but the first graph is linear with a positive slope while the second graph curves upward, becoming steeper and steeper.