Question

instructions: analyze the position-time graph of an object shown in figure 3.10. then, answer the following questions.

1. what is the speed of the object from 0 to 3 seconds?
2. what happened to the motion of the object from 3.0 to 4.0 seconds in terms of direction and velocity? by how much is the velocity of the object?
3. for the last five seconds (from 4.0 to 9.0 seconds), what happened to the object’s motion?

graph: position vs. time graph of an object, position (meters) on y - axis (0 - 20), time (seconds) on x - axis (0 - 12)

Explanation:

Question 1:

Step 1: Recall the formula for speed from a position - time graph

Speed is calculated as the slope of the position - time graph, and the formula for slope (speed) when dealing with a position - time graph is $v=\frac{\Delta x}{\Delta t}$, where $\Delta x$ is the change in position and $\Delta t$ is the change in time.

Step 2: Determine the values of $\Delta x$ and $\Delta t$ from 0 to 3 seconds

From the graph, at $t = 0$ seconds, the position $x_1=0$ meters. At $t = 3$ seconds, the position $x_2 = 12$ meters (assuming the graph shows that at $t = 3$ s, the position is 12 m). The change in time $\Delta t=t_2 - t_1=3 - 0 = 3$ seconds. The change in position $\Delta x=x_2 - x_1=12 - 0 = 12$ meters.

Step 3: Calculate the speed

Using the formula $v=\frac{\Delta x}{\Delta t}$, we substitute $\Delta x = 12$ m and $\Delta t=3$ s. So $v=\frac{12}{3}=4$ m/s.

Step 1: Analyze the direction and velocity from 3.0 to 4.0 seconds

In a position - time graph, the slope of the line gives the velocity. If the slope is negative, the object is moving in the opposite direction (towards the origin or in the negative direction of position). The formula for velocity is still $v=\frac{\Delta x}{\Delta t}$.

Step 2: Determine the values of $\Delta x$ and $\Delta t$ from 3.0 to 4.0 seconds

At $t = 3$ seconds, the position $x_1 = 12$ meters. At $t = 4$ seconds, the position $x_2=0$ meters. The change in time $\Delta t=t_2 - t_1=4 - 3 = 1$ second. The change in position $\Delta x=x_2 - x_1=0 - 12=- 12$ meters.

Step 3: Calculate the velocity

Using the formula $v=\frac{\Delta x}{\Delta t}$, we substitute $\Delta x=- 12$ m and $\Delta t = 1$ s. So $v=\frac{-12}{1}=- 12$ m/s. The negative sign indicates that the object is moving in the opposite direction (reverse direction) compared to its motion from 0 to 3 seconds. The magnitude of the velocity is 12 m/s.

Step 1: Analyze the position - time graph from 4.0 to 9.0 seconds

In a position - time graph, if the position of the object does not change with time (the graph is a horizontal line), the object is at rest (its velocity is zero).

Step 2: Check the position values from 4.0 to 9.0 seconds

From the graph, at $t = 4$ seconds, the position is 0 meters. For all the times from $t = 4$ seconds to $t = 9$ seconds, the position of the object remains 0 meters. This means that the change in position $\Delta x = 0$ for any $\Delta t$ in this time interval.

Step 3: Determine the motion

Using the formula for velocity $v=\frac{\Delta x}{\Delta t}$, since $\Delta x = 0$, the velocity $v = 0$. So the object is at rest (stationary) from 4.0 to 9.0 seconds.

Answer:

The speed of the object from 0 to 3 seconds is 4 m/s.

Question 2: