Time (s): 0 1 2 3 4 5 7 8 9 10
Δ X (m/s)
Position:
0 3 7 12 17 24 22 17 14 14
1. Graph the table above (include title, label the independent (x) and dependent variable (y), use the proper scale)
2. Identify on your scale the positive acceleration and the negative acceleration.
3. For how many seconds was the motion of ball traveling at an increase in velocity (positive acceleration)?
3. At which second did themotion of the ball start to show a decrease in velocity (negative acceleration)?
4. At any point did the motion of the ball travel at a constant acceleration? When?
5. What occured in the 9th and the 10th second with the motion of the ball?
1. To graph the table, we'll plot the time (s) on the x-axis and the position (ΔX in m/s) on the y-axis.
The title for the graph could be "Position vs Time Graph".
On the x-axis (independent variable), we will mark the time points: 0, 1, 2, 3, 4, 5, 7, 8, 9, 10.
On the y-axis (dependent variable), we will mark the position points: 0, 3, 7, 12, 17, 24, 22, 17, 14, 14.
Make sure to choose a suitable scale for both axes so that all the points can be properly represented on the graph.
2. To identify positive acceleration and negative acceleration, we need to look at the change in velocity (ΔX) values.
Positive acceleration occurs when the change in velocity is increasing (getting larger), and negative acceleration occurs when the change in velocity is decreasing (getting smaller).
Looking at the table, we can see that the change in velocity is increasing until the 5th second (from 0 to 17 m/s). This represents positive acceleration.
After the 5th second, the change in velocity starts to decrease (from 24 to 14 m/s). This represents negative acceleration.
3. We need to determine for how many seconds the ball was traveling at an increase in velocity (positive acceleration).
From the table, we can see that the change in velocity (ΔX) is positive (increasing) until the 5th second.
So, the motion of the ball was traveling at an increase in velocity for 5 seconds.
4. To find the second at which the motion of the ball starts to show a decrease in velocity (negative acceleration), we can look at the table.
At the 6th second (time = 6), the change in velocity is the highest (from 24 to 22 m/s) and then starts to decrease afterward.
So, the motion of the ball starts to show a decrease in velocity (negative acceleration) at the 6th second.
5. To determine if the motion of the ball traveled at a constant acceleration, we need to check if the change in velocity (ΔX) is the same for any time interval.
From the table, we can observe that the change in velocity is not constant, as it varies for each time interval. Therefore, the motion of the ball did not travel at a constant acceleration.
6. Finally, in the 9th and 10th seconds, we can see that the motion of the ball has the same position value of 14 m/s. This means that in the 9th and 10th seconds, the motion of the ball remained at the same position without any change in velocity.