A car traveling at 4.0 m/s has a constant acceleration of 2.0 m/s² in the same direction as the velocity. After 3.0 seconds, the distance traveled is
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.
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![**Problem 12:**
A car traveling at 4.0 m/s has a constant acceleration of 2.0 m/s² in the same direction as the velocity. After 3.0 seconds, the distance traveled is:
### Solution:
To find the distance traveled, you can use the kinematic equation:
\[ s = ut + \frac{1}{2}at^2 \]
Where:
- \( s \) is the distance traveled,
- \( u \) is the initial velocity (\( 4.0 \, \text{m/s} \)),
- \( a \) is the acceleration (\( 2.0 \, \text{m/s}^2 \)),
- \( t \) is the time (\( 3.0 \, \text{s} \)).
Plugging in the values:
\[ s = (4.0 \, \text{m/s})(3.0 \, \text{s}) + \frac{1}{2}(2.0 \, \text{m/s}^2)(3.0 \, \text{s})^2 \]
\[ s = 12.0 \, \text{m} + \frac{1}{2}(2.0)(9.0) \]
\[ s = 12.0 \, \text{m} + 9.0 \, \text{m} \]
\[ s = 21.0 \, \text{m} \]
Therefore, the car travels a distance of 21.0 meters in 3.0 seconds.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fffd53d82-b6a8-4210-805b-6a3c6ec01562%2F181771b6-84ec-4426-a588-1c78532c852f%2Fppgmmn_processed.jpeg&w=3840&q=75)
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