A student begins at rest and then walks north with an average speed of v1 = 0.85 m/s. The student then turns south and walks with an average speed v2 = 0.45 m/s.
(a) What was the student's overall average velocity vavg, x, in m/s, for the trip assuming the student spent equal times at speeds v1 and v2?
Well, it seems like this student had a pretty balanced trip, like a true Olympic gymnast on the beam! So, let's calculate their overall average velocity, shall we?
To find the overall average velocity, we need to consider both the magnitude and direction of the velocities involved. Since the student walks both north and south, we can assume that the overall average velocity will be in the southern direction. Let's see!
The average speed v1 for the northward walk is 0.85 m/s, while the average speed v2 for the southward walk is 0.45 m/s. Since the student spent equal times at these two speeds, we can consider their magnitudes to remain the same.
Now, to calculate the overall average velocity, we need to consider the displacement. Since the student walks north and then south, their displacement will be in the opposite direction of the displacement during the northward walk.
Let's say the displacement during the northward walk is d1, and the displacement during the southward walk is d2. Since the magnitudes of the velocities remain the same, we can use the formula vavg = (d1 + d2)/(t1 + t2), where t1 and t2 are the times spent at each velocity.
Since the time spent at each velocity is equal, we can simplify the formula to vavg = (d1 + d2)/2.
Now, the student walks at the same speed for each leg of the journey, so we can use vavg = (v1 + v2)/2.
Plugging in the values, we get vavg = (0.85 m/s + 0.45 m/s)/2 = 1.3 m/s.
Therefore, the student's overall average velocity, vavg, is 1.3 m/s in the southern direction.
Looks like this student turned their trip into a Southern Comfort party!
To find the overall average velocity, we need to calculate the average of the velocities for both segments. The first segment has a velocity of v1 = 0.85 m/s, and the second segment has a velocity of v2 = 0.45 m/s.
The average velocity is given by the formula:
vavg = (v1 + v2) / 2
Substituting the given values, we have:
vavg = (0.85 m/s + 0.45 m/s) / 2
Simplifying the expression, we get:
vavg = 1.3 m/s / 2
vavg = 0.65 m/s
Therefore, the student's overall average velocity is 0.65 m/s.
To find the overall average velocity (vavg), we'll use the concept of average velocity. Average velocity is defined as the total displacement of an object divided by the total time taken.
In this scenario, the student walks north and then turns south, so the displacement in the north direction cancels out the displacement in the south direction. Thus, the total displacement is zero.
Now, we need to find the time taken for each segment of the trip. Let's assume the time spent at each speed (v1 and v2) is equal. Let's call this time t.
The distance covered at a constant speed can be calculated using the formula: distance = speed × time.
For the first part of the trip, when the student walks north with a speed of v1 = 0.85 m/s, the distance covered is 0.85t.
For the second part of the trip, when the student walks south with a speed of v2 = 0.45 m/s, the distance covered is 0.45t.
Since the total displacement is zero, the distances covered in the north and south directions should be equal. Therefore, 0.85t = 0.45t.
Simplifying the equation, we get:
0.85t = 0.45t
0.4t = 0
Since the second part of the equation is equal to zero, we get t = 0.
This means that the student spends zero time at each speed, which is not possible. Therefore, this scenario is not physically realistic, and we cannot find the overall average velocity without knowing the total time or a different set of assumptions.
So, in conclusion, we can't determine the student's overall average velocity in this specific scenario.