Q3. Notice that the rate at which "IT" changes is the greatest at first, i.e., the slope at "A" (shown below) is greater than the slope at "B". Explain why in this space:
Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
I need help with Q3 and Q4. Thanks.
![**Q3**. Notice that the rate at which “I” changes is the greatest at first, i.e., the slope at “A” (shown below) is greater than the slope at “B”. Explain why in this space:
*Diagram:*
The graph shows time \(t\) on the x-axis and current \(I\) on the y-axis, with points “A” and “B” indicating steep and less steep slopes, respectively.
*Note:* If you are mathematically inclined, you can see this using Kirchhoff’s Voltage Rule for this circuit:
\[ V = L \frac{\Delta I}{\Delta t} + I R \]
*Diagram:*
A circuit diagram showing components like an inductor (\(L\)) and a resistor (\(R\)) with constant labels.
By looking at the I versus t graph, you might have guessed that an exponential function “fits” it. If so, you’re right! Here it is:
\[ I = I_{\text{max}} \left(1 - e^{-t/\tau}\right) \]
*Equation 2*
**Q4**. Using Equation 1 and Equation 2, determine “I” in terms of “Imax” at the times given in the table below. The third one is done for you: At \(t = \tau\)
\[ I = I_{\text{max}} \left(1 - e^{-t/\tau}\right) \]
\[ = I_{\text{max}} \left(1 - e^{-1}\right) \]
\[ \approx I_{\text{max}} \left(1 - 0.37\right) \]
\[ = I_{\text{max}} (0.63) \]
Insert your answers in the table below.
*Graph:*
The graph emphasizes \(t = \tau\) with current \( I = 0.63 I_{\text{max}} \) and prompts sketching results at \( t = 0 \), \( t = 0.5\tau \), and \( t = 3\tau \).
*Table:*
\[
\begin{array}{|c|c|}
\hline
\text{Time} & \text{Current} \\
\hline
t = 0 & \\
t = 0.5\tau & \\
t = \tau](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F43da8127-b8b5-4d0a-8317-0b1da88f2197%2Ff9138623-dbf2-45c0-bf5c-042daf295a33%2Fydivpxc_processed.png&w=3840&q=75)
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