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 car is traveling at a speed of 95.0 km/hr. The driver applies the brakes and the car comes to a stop over a distance of 60.0 meters.
**Question:**
What is the car's deceleration?
**Solution Explanation:**
To find the deceleration, we can use the formula:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity (0 m/s, since the car stops)
- \( u \) is the initial velocity (95.0 km/hr converted to m/s)
- \( a \) is the acceleration (deceleration value to be found)
- \( s \) is the stopping distance (60.0 m)
First, convert 95.0 km/hr to meters per second:
\[ 95.0 \, \text{km/hr} = \frac{95.0 \times 1000}{3600} \, \text{m/s} \]
Next, we rearrange the formula to solve for \( a \):
\[ 0 = u^2 + 2(-a)s \]
\[ a = \frac{-u^2}{2s} \]
Plug in the values to find the deceleration.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F27e7e80b-03df-453b-a555-8f500263e07b%2Fa9a0bf21-4972-4c0c-897f-879e8a42c233%2Fhra9pr_processed.jpeg&w=3840&q=75)
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