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.
![**Problem Statement:**
A girl rides her bike \(4.50\) blocks west, \(3.00\) blocks north, and then \(7.25\) blocks east.
**Questions:**
1. **What is the girl's displacement?**
(Give the magnitude of your answer in terms of blocks and the direction in degrees north of east.)
\[ \_\_\_\_\_ \text{ blocks at } \_\_\_\_\_ ^\circ \text{ north of east} \]
2. **What total distance (in blocks) does the girl ride?**
\[ \_\_\_\_\_ \text{ blocks} \]
---
**Explanation for Educators:**
When solving for displacement, one must first understand that displacement is a vector quantity that includes both magnitude and direction. The given problem involves breaking the girl's journey into vector components and then finding the resultant displacement vector.
1. **Finding Displacement:**
- The girl rides \(4.50\) blocks west (negative x-direction).
- She then rides \(3.00\) blocks north (positive y-direction).
- Finally, she rides \(7.25\) blocks east (positive x-direction).
To find the net displacement in the x-direction:
\[
x_{\text{net}} = 7.25 \text{ blocks (east)} - 4.50 \text{ blocks (west)} = 2.75 \text{ blocks}
\]
The net displacement in the y-direction:
\[
y_{\text{net}} = 3.00 \text{ blocks}
\]
Using the Pythagorean Theorem to find the displacement magnitude:
\[
\text{Magnitude} = \sqrt{{x_{\text{net}}}^2 + {y_{\text{net}}}^2} = \sqrt{{2.75}^2 + {3.00}^2}
\]
The formula for the direction (angle) θ north of east:
\[
\theta = \tan^{-1}\left(\frac{y_{\text{net}}}{x_{\text{net}}}\right)
\]
2. **Total Distance Traveled:**
- Sum of individual distances traveled, regardless of direction:
\[
\text{Total Distance} = 4.50](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb82cdd93-87a8-4004-913f-cae9a7e1e525%2F4e36c8f7-9802-4aaf-bb06-d031f9d25252%2F8zykrju_processed.jpeg&w=3840&q=75)
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