Figure 2-16 gives the velocity of a particle moving on an x axis. What are (a) the initial and (b) the final directions of travel? (c) Does the particle stop momentarily? (d) Is the acceleration positive or negative? (e) Is it constant or varying? Figure 2-16 Question 1.
Figure 2-16 gives the velocity of a particle moving on an x axis. What are (a) the initial and (b) the final directions of travel? (c) Does the particle stop momentarily? (d) Is the acceleration positive or negative? (e) Is it constant or varying? Figure 2-16 Question 1.
Figure 2-16 gives the velocity of a particle moving on an x axis. What are (a) the initial and (b) the final directions of travel? (c) Does the particle stop momentarily? (d) Is the acceleration positive or negative? (e) Is it constant or varying?
Figure 2-16 Question 1.
Expert Solution & Answer
To determine
To find:
a) Initial direction of travel.
b) Final direction of travel.
c) Does the particle stop momentarily?
d) Is the acceleration positive or negative?
e) Is the acceleration constant or varying?
Answer to Problem 1Q
Solution:
a) Initial direction of travel is negative.
b) Final direction of travel is positive.
c) The particle stops momentarily.
d) The acceleration is positive.
e) The acceleration is constant.
Explanation of Solution
1) Concept:
After reading the graph, we can say that the Initial direction of travel is negative. As compared to that final direction of travel is positive.
As for an instance the velocity of the particle is zero, it indicates that the particle stops momentarily.
From the graph, we can say that the acceleration is positive and constant as the graph is a straight line and ascending in positive direction.
2) Given:
Graph in Fig 2.16.
3) Calculations:
a) When we look at the graph, it is starting below the x-axis, which states that the initial direction of travel would be negative.
b) When we look at the graph, it is ending above the x-axis, which states that the final direction of travel would be positive.
c) When we say it stops momentarily, it means the velocity of the particle is zero for a moment. When we look at the graph, the velocity is zero for one point, so we can say that the particle stops momentarily.
d) Now as the graph is moving from negative velocity to positive velocity, it means the velocity is increasing. From the above statement, we can say that the acceleration is positive.
e) As the graph is a straight line, the acceleration is constant.
Conclusion:
We have been given the graph of velocity vs time. From this graph, we can find the acceleration of the object by finding the slope of the graph. Looking at the nature of the graph, we can also conclude the direction of the acceleration and velocity at any given point.
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Can someone help me answer this physics 2 questions. Thank you.
Four capacitors are connected as shown in the figure below. (Let C = 12.0 μF.)
a
C
3.00 με
Hh.
6.00 με
20.0 με
HE
(a) Find the equivalent capacitance between points a and b.
5.92
HF
(b) Calculate the charge on each capacitor, taking AV ab = 16.0 V.
20.0 uF capacitor 94.7
6.00 uF capacitor 67.6
32.14
3.00 µF capacitor
capacitor C
☑
με
με
The 3 µF and 12.0 uF capacitors are in series and that combination is in parallel with the 6 μF capacitor. What quantity is the same for capacitors in parallel? μC
32.14
☑
You are correct that the charge on this capacitor will be the same as the charge on the 3 μF capacitor. μC
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