The figure shows the y-position (in blue) of a particle versus time.
(a) What is the average velocity of the particle during the time interval
t = 1.50 s
to
t = 4.50 s?
(Express your answer in vector form.)
Incorrect: Your answer is incorrect.
webMathematica generated answer key m/s
(b) Using the tangent to the curve (shown as the orange line in the figure), what is the instantaneous velocity of the particle at
t = 1.50 s?
(Express your answer in vector form.)
m/s
(c) At what time is the velocity of the particle equal to zero?
s
*For a) the point for 1.50s is (1.50s, 11m) and for 4.50s is (4.50s,17.5m)
*Plug in thee average velocity formula: Vav= delta y over delta t which is(17.5m-11m)/(4.50s-1.50s)=2.17jm/s
*For b) you need another point on the orange line, let chose point(3s,20m)
* Plug in the instaneous formula: S= delta y over delta t which is (20m-11m)/(3s-1.50s)=6.00m/s
*For c) Look at the highest point on the graph. In this case it is 3.50s.
These questions cannot be answered with the diagram that they are based on.
(a) To find the average velocity of the particle during the time interval from t = 1.50 s to t = 4.50 s, we need to calculate the displacement of the particle during this time interval and divide it by the time elapsed.
The displacement is the change in position of the particle. Since the position is given as a function of time, we can find the change in position by subtracting the initial position from the final position.
Given that the y-position of the particle is shown in blue, we need to find the y-coordinate at t = 1.50 s and t = 4.50 s. Let's denote the position at t = 1.50 s as y1 and the position at t = 4.50 s as y2.
Then, the displacement in the y-direction is:
Δy = y2 - y1
To express the average velocity in vector form, we need to divide the displacement by the time interval:
Average velocity = Δy / Δt
Given the time interval Δt = 4.50 s - 1.50 s = 3.00 s, we can calculate the average velocity by substituting the values:
Average velocity = (y2 - y1) / Δt
Unfortunately, we do not have the specific values for y1 and y2, so we cannot provide the exact vector form of the average velocity. You would need to provide the specific values for y1 and y2 to calculate the average velocity.
(b) To find the instantaneous velocity of the particle at t = 1.50 s, we can use the tangent line to the curve at that point. The tangent line represents the instantaneous velocity at that specific time.
By drawing a tangent line to the curve at t = 1.50 s, we can determine its slope. The slope of the tangent line represents the instantaneous velocity.
Once again, we do not have the specific values for the slope or the angle of the tangent line, so we cannot provide the exact vector form of the instantaneous velocity. You would need to provide additional information or equations to determine the slope and calculate the instantaneous velocity.
(c) To find the time at which the velocity of the particle is equal to zero, we need to identify the point on the graph where the tangent line is horizontal or flat.
However, without a specific graph or data points, we cannot determine the exact time at which the velocity is zero. You would need to provide more information or equations to analyze the graph further and find the time when the velocity is zero.
To answer these questions, we need the figure that shows the y-position of the particle versus time. However, I can explain how you can solve these types of problems.
(a) To find the average velocity of the particle during the time interval from t = 1.50 s to t = 4.50 s, you need to calculate the change in position (Δy) and divide it by the change in time (Δt) within that interval. The average velocity can be represented as:
Average velocity = Δy / Δt
To calculate Δy, find the difference between the initial y-position at t = 1.50 s and the y-position at t = 4.50 s. Similarly, calculate Δt by subtracting the initial time from the final time.
(b) To find the instantaneous velocity at t = 1.50 s, you can use the tangent line to the curve shown in the figure. The tangent line represents the velocity of the particle at any given time. Determine the slope of the tangent line at t = 1.50 s, which corresponds to the instantaneous velocity at that moment. You can calculate the slope using the rise over run method:
Instantaneous velocity = Slope of tangent line at t = 1.50 s
(c) To find the time at which the velocity of the particle is zero, you need to locate the point on the curve where the tangent line is horizontal (having a slope of zero). This point indicates that the particle is momentarily at rest, thus having zero velocity. Find the corresponding time on the x-axis of the figure.
Once you provide the figure, I can help you solve these specific questions.