A car starts 200 m west of the town square and moves with a constant velocity of 15 m/s toward the east. Draw a graph that represents the motion of the car a. Where will the car be 10 minutes later? When will the car reach the town square? b.
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.
Use sketches on velocity vs time and acceleration vs time graphs to make your case.
![### Problem Statement:
A car starts 200 meters west of the town square and moves with a constant velocity of 15 meters per second towards the east.
1. **Draw a graph that represents the motion of the car.**
2. **Answer the following questions:**
a. Where will the car be 10 minutes later?
b. When will the car reach the town square?
### Explanation:
To solve this problem, we need to:
1. **Graphing the Motion of the Car:**
- The x-axis of the graph will represent time (in seconds), starting from 0.
- The y-axis of the graph will represent the position of the car relative to the town square (in meters), starting from -200 meters (200 meters west of the town square).
- Since the car has a constant velocity, the position of the car changes linearly over time. The slope of this line is equal to the car’s velocity, which is 15 meters per second.
2. **Solving the Questions:**
a. **Where will the car be 10 minutes later?**
- First, convert 10 minutes to seconds:
\[
10 \text{ minutes} \times 60 \text{ seconds/minute} = 600 \text{ seconds}
\]
- Using the equation of motion (position \(s\) as a function of time \(t\)):
\[
s(t) = s_0 + vt
\]
Where:
- \(s(t)\) is the position at time \(t\).
- \(s_0\) is the initial position (which is -200 meters).
- \(v\) is the velocity (which is 15 meters/second).
- Substituting the values:
\[
s(600) = -200 + 15 \times 600
\]
\[
s(600) = -200 + 9000
\]
\[
s(600) = 8800 \text{ meters}
\]
- Therefore, the car will be 8800 meters east of the town square after 10 minutes.
b. **When will the car reach the town square?**
- To find when the car reaches the position of 0 meters (the town square), set \(s(t)\) to](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b60de5f-8fb0-4e3b-8258-2de6c1a93a8c%2F95318210-8920-40a1-9d78-086337b931d1%2Flhzqay_processed.png&w=3840&q=75)
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