Question

1. suppose your speed at the final mark was 0.95 m/s. calculate your average rate of change from the start through the final segment of the course.

Explanation:

1. Explanation:

• Step 1: Recall the formula for average rate of change of speed
• The average rate of change of a quantity \(y\) with respect to \(x\) over an interval \([a,b]\) is given by \(\frac{\Delta y}{\Delta x}=\frac{y(b)-y(a)}{b - a}\). In the context of speed, if we assume the initial - speed \(v_0 = 0\) (since we start from rest, not given otherwise) and the final speed \(v_f=0.95\ m/s\), and we consider the time interval from the start (\(t = 0\)) to the end (\(t=t_f\)). The average rate of change of speed (which is also the average acceleration if the initial speed is \(0\)) over the entire course is \(\frac{v_f - v_0}{t_f-0}\). But if we just want the average rate of change of speed from start to the final mark and assume the start speed \(v_0 = 0\), the formula simplifies to \(\frac{v_f - v_0}{t}\). Since \(v_0 = 0\) and \(v_f=0.95\ m/s\), and if we assume the time taken from start to the final mark is \(t\) seconds, the average rate of change of speed \(r=\frac{0.95 - 0}{t}\). However, if we assume the motion starts from rest (\(v_0 = 0\)) and we are not given any information about time, and we consider the average rate of change of speed in terms of just the initial and final values of speed, the average rate of change of speed from start to the final mark is \(\frac{\text{Final Speed}-\text{Initial Speed}}{\text{Time}}\). If we assume the start - speed \(v_0 = 0\) and the final speed \(v_f = 0.95\ m/s\), and for simplicity, if we assume the time interval from start to finish is \(t = 1\) second (since no time information is given), the average rate of change of speed is \(\frac{0.95-0}{1}=0.95\ m/s^2\).

1. Answer:

• \(0.95\ m/s^2\) (assuming the start - speed is \(0\ m/s\) and the time interval from start to the final mark is \(1\) second)

Answer:

1. Explanation:

• Step 1: Recall the formula for average rate of change of speed
• The average rate of change of a quantity \(y\) with respect to \(x\) over an interval \([a,b]\) is given by \(\frac{\Delta y}{\Delta x}=\frac{y(b)-y(a)}{b - a}\). In the context of speed, if we assume the initial - speed \(v_0 = 0\) (since we start from rest, not given otherwise) and the final speed \(v_f=0.95\ m/s\), and we consider the time interval from the start (\(t = 0\)) to the end (\(t=t_f\)). The average rate of change of speed (which is also the average acceleration if the initial speed is \(0\)) over the entire course is \(\frac{v_f - v_0}{t_f-0}\). But if we just want the average rate of change of speed from start to the final mark and assume the start speed \(v_0 = 0\), the formula simplifies to \(\frac{v_f - v_0}{t}\). Since \(v_0 = 0\) and \(v_f=0.95\ m/s\), and if we assume the time taken from start to the final mark is \(t\) seconds, the average rate of change of speed \(r=\frac{0.95 - 0}{t}\). However, if we assume the motion starts from rest (\(v_0 = 0\)) and we are not given any information about time, and we consider the average rate of change of speed in terms of just the initial and final values of speed, the average rate of change of speed from start to the final mark is \(\frac{\text{Final Speed}-\text{Initial Speed}}{\text{Time}}\). If we assume the start - speed \(v_0 = 0\) and the final speed \(v_f = 0.95\ m/s\), and for simplicity, if we assume the time interval from start to finish is \(t = 1\) second (since no time information is given), the average rate of change of speed is \(\frac{0.95-0}{1}=0.95\ m/s^2\).

1. Answer:

• \(0.95\ m/s^2\) (assuming the start - speed is \(0\ m/s\) and the time interval from start to the final mark is \(1\) second)