Question

use the following position - time graph of a squirrel running along a clothesline to answer questions 5 - 6.

1. what is the squirrels displacement at time t = 3.0s?
2. what is the squirrels average velocity during the time interval between 0.0 s and 3.0 s?
3. explain what acceleration is:
4. a ball initially at rest rolls down a hill and has an acceleration of 3.3 m/s². if it accelerates for 7.5 s, how far will it move during this time?

Explanation:

Question 5:

Step1: Understand Displacement from Graph

Displacement is the position at a given time. From the position - time graph, at \(t = 3.0\space s\), we look at the y - axis (position) value. The graph shows that at \(t=3.0\space s\), the position (displacement) is \(- 1.0\space m\) (assuming the y - axis is position in meters). Wait, maybe I misread. Wait, the graph: from \(t = 2.0\) to \(t = 3.0\), the position goes to \(-1.0\)? Wait, let's re - examine. The graph has time on x - axis (0 to 5 s) and position on y - axis. At \(t = 3.0\space s\), the position is \(-1.0\space m\)? Wait, no, maybe the user's initial note had a mistake, but let's go by the graph. Wait, the first part: from \(t = 0\) to \(t = 1.0\), it goes up to 4.0? No, wait the y - axis is from - 2.0 to 4.0. Wait, at \(t = 0\), position is 0. At \(t = 1.0\), position is 4.0? Then at \(t = 2.0\), position is 0. Then from \(t = 2.0\) to \(t = 3.0\), it goes down to - 1.0? Wait, maybe the correct position at \(t = 3.0\space s\) is \(-1.0\space m\)? But the user's initial writing had - 2m, maybe a typo. Anyway, let's proceed.

Step2: Confirm Position at \(t = 3.0\space s\)

Looking at the position - time graph, at \(t=3.0\space s\), the y - coordinate (position) is \(- 1.0\space m\)? Wait, no, maybe the graph's scale: the vertical axis (position) has marks at - 2.0, - 1.0, 0, 1.0, 2.0, 3.0, 4.0. At \(t = 3.0\space s\), the point is at \(y=-1.0\space m\)? Or maybe \(-2.0\)? Wait, the graph after \(t = 2.0\) goes down to a minimum at \(t = 3.0\)? Wait, the x - axis is time (s): 0,1,2,3,4,5. The y - axis is position (m): - 2, - 1, 0,1,2,3,4. At \(t = 3.0\space s\), the position is \(-1.0\space m\)? Or maybe the user's initial answer was wrong. Anyway, the displacement at \(t = 3.0\space s\) is the position at that time, since displacement from origin (t = 0, position = 0) is position at t. So if at \(t = 3.0\space s\), position is \(-1.0\space m\), then displacement is \(-1.0\space m\). But maybe the graph is different. Wait, maybe the first triangle: from (0,0) to (1,4) to (2,0), then from (2,0) to (3, - 1) to (4, - 1) to (5,0). So at \(t = 3.0\space s\), position is \(-1.0\space m\).

Step1: Recall Average Velocity Formula

Average velocity \(v_{avg}=\frac{\Delta x}{\Delta t}\), where \(\Delta x=x_f - x_i\) (displacement) and \(\Delta t=t_f - t_i\).

Step2: Find Initial and Final Positions

At \(t_i = 0.0\space s\), \(x_i = 0\space m\) (from the graph, at \(t = 0\), position is 0). At \(t_f=3.0\space s\), from question 5, if \(x_f=-1.0\space m\) (or - 2.0m), then \(\Delta x=x_f - x_i=-1.0 - 0=-1.0\space m\) (or \(-2.0 - 0=-2.0\space m\)). \(\Delta t=3.0 - 0.0 = 3.0\space s\).

Step3: Calculate Average Velocity

If \(x_f=-1.0\space m\), then \(v_{avg}=\frac{-1.0\space m}{3.0\space s}\approx - 0.33\space m/s\). If \(x_f=-2.0\space m\), then \(v_{avg}=\frac{-2.0\space m}{3.0\space s}\approx - 0.67\space m/s\). But let's check the graph again. Wait, at \(t = 3.0\space s\), the position is \(-1.0\space m\)? Or maybe the final position at \(t = 3.0\space s\) is \(-1.0\space m\). So \(\Delta x=-1.0\space m\), \(\Delta t = 3.0\space s\), so \(v_{avg}=\frac{-1.0}{3.0}\approx - 0.33\space m/s\).

Acceleration is the rate of change of velocity with respect to time. Mathematically, \(a=\frac{\Delta v}{\Delta t}\), where \(\Delta v\) is the change in velocity and \(\Delta t\) is the change in time. It can also be thought of as how quickly an object's speed or direction of motion changes. If velocity is constant, acceleration is zero. If velocity increases (in speed or changes direction to speed up), acceleration is positive; if velocity decreases (deceleration, which is negative acceleration), acceleration is negative.

Answer:

The squirrel's displacement at \(t = 3.0\space s\) is \(\boldsymbol{-1.0\space m}\) (or if the graph shows \(-2.0\space m\), then \(\boldsymbol{-2.0\space m}\) depending on the graph's exact plot).

Question 6: