Question

(01.10 mc) an object travels east from the starting position of (0, 0). a position versus time graph is shown below. position vs. time graph (description: position on y-axis, time on x-axis; initial linear increase then horizontal) this object was moving at a velocity of 1.0 m/s east at the end of 4.0 seconds. determine the average and instantaneous velocities in m/s at 8.0 seconds.
○ a average = 1.0 m/s east, instantaneous = 4.0 m/s east
○ b average = 4.0 m/s east, instantaneous = 8.0 m/s east
○ c average = 1.0 m/s east, instantaneous = 0 m/s
○ d average = 0.50 m/s east, instantaneous = 0 m/s

Explanation:

Step1: Recall velocity formulas

Average velocity is given by $\frac{\Delta x}{\Delta t}$, where $\Delta x$ is displacement and $\Delta t$ is time interval. Instantaneous velocity is the slope of the position - time graph at a specific time.

Step2: Determine displacement and time for average velocity

The object starts at $(0,0)$ and at $t = 8.0\ s$, from the graph, the position $x = 4.0\ m$ (assuming the y - axis is position in meters, since at $t = 4.0\ s$, position is $4.0\ m$ and then it remains constant). The time interval $\Delta t=8.0 - 0=8.0\ s$, and displacement $\Delta x = 4.0-0 = 4.0\ m$? Wait, no, wait. Wait, at $t = 4\ s$, position is $4\ m$, and then it stays at $4\ m$ until $t = 8\ s$. Wait, initial position $x_0 = 0\ m$ at $t_0 = 0\ s$, final position $x_f=4\ m$ at $t_f = 8\ s$. So average velocity $v_{avg}=\frac{x_f - x_0}{t_f - t_0}=\frac{4 - 0}{8 - 0}=\frac{4}{8}=0.5\ m/s$? Wait, no, wait the problem says "at the end of 4.0 seconds" velocity is $1.0\ m/s$ east. Wait, let's re - examine the graph. From $t = 0$ to $t = 4\ s$, the position - time graph is a straight line. The slope from $t = 0$ to $t = 4\ s$: at $t = 0$, $x = 0$; at $t = 4\ s$, $x = 4\ m$. So slope (velocity) is $\frac{4 - 0}{4 - 0}=1\ m/s$, which matches the given information. Then from $t = 4\ s$ to $t = 8\ s$, the position remains constant (horizontal line), so the slope (instantaneous velocity) is $0\ m/s$ (since $\Delta x = 0$ for $\Delta t>0$).

Now, for average velocity at $t = 8\ s$: total displacement $\Delta x$ from $t = 0$ to $t = 8\ s$ is $4\ m$ (since at $t = 8\ s$, position is $4\ m$), total time $\Delta t = 8\ s$. So $v_{avg}=\frac{4 - 0}{8 - 0}=0.5\ m/s$ east.

Instantaneous velocity at $t = 8\ s$: the graph is horizontal from $t = 4\ s$ to $t = 8\ s$, so the slope (instantaneous velocity) is $0\ m/s$.

Now let's check the options:

Option a: Average = 1.0 m/s (wrong, we calculated 0.5), instantaneous = 4.0 m/s (wrong, slope is 0).

Option b: Average = 4.0 m/s (wrong), instantaneous = 8.0 m/s (wrong).

Option c: Average = 1.0 m/s (wrong), instantaneous = 0 m/s (instantaneous is correct, but average is wrong).

Option d: Average = 0.50 m/s east, instantaneous = 0 m/s. This matches our calculations.

Answer:

d. Average = 0.50 m/s east; instantaneous = 0 m/s