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Physics Formulas

Physics Kinematics Formulas

Kinematics is the popular branch of Physics, (in particular classical mechanics), which describes the motion of objects. Also, it covers the individual object or systems of groups of objects. It does not take the reference of causes of the motion i.e., forces, etc. This article will explain the Physics kinematics formulas with examples. Let us learn it in detail here.

Concept of Kinematics

The goal of kinematics is to develop the sophisticated mental models that serve to describe and hence explain the motion of real-world objects. There are many physical quantities associated with the motion of the objects. Some of these terms are displacement or distance, velocity or speed, acceleration, and the time. Knowledge of each of these quantities will provide descriptive information about any object in motion.

Physics Kinematics Formulas

Source: en.wikipedia.org

These equations can be used for any motion which can be described as being either a constant velocity motion or a constant acceleration motion. They cannot be used over any time period during which the acceleration is changing. Each of the kinematic formulas includes four variables. If the three of them variables are known, then the value of the fourth variable can be computed. Thus the kinematic equations provide a useful method of predicting information about an object’s motion.

Kinematics Formula is mainly about the motion of bodies at some points without considering the cause through which motion is happening.

The Formula for Kinematics:

There are mainly four kinematic equations, which are given as follows:

\(v=v_{0}+at\)

\(\Delta x=(\frac{v+v_{0}}{2})t\)

\(\Delta x=v_{0}t+\frac{1}{2}at^{2}\)

\(v^{2}=v_{0}^{2}+2a\Delta x\)

Where,

\(v_{0}\) initial velocity
v velocity at a certain time t.
\(\Delta x\) the displacement/distance travelled, due to change in the position
a Acceleration
t Time

Solved Examples for Physics Kinematics Formulas

Q.1: A boy is riding a bike with the initial velocity of \(2 ms^{-1}\). He reaches his destination after 3 sec having a final velocity of \(10 ms^{-1}\). Determine his acceleration due to motion.

Solution: Given parameters are as follows,

  • Initial Velocity \(v_{o} = 2 ms^{-1}\\\)
  • Final velocity \(v = 10 ms^{-1}\\\)
  • Time period t = 3 sec

Now, to find the acceleration we will use the formula:

\(v=v_{0}+at\)

Acceleration will be

\(a = \frac{v – v_{0}}{t} \\\)

\(= \frac{10-2}{3}\\\)

\(= 2.76 ms^{-2}\\\)

Thus acceleration will be \(2.76 ms^{-2}.\)

Q.2: If a vehicle accelerates with the acceleration of \(3.0 ms^{-2}\) starting from a complete stop. Then, how long will it take to go 3000 m? Use Physics Kinematics Formulas.

Solution: Given here,

The acceleration, \(a = 3.0 ms^{-2},\)

and the displacement, \(\Delta x = 3000 m.\)

The vehicle was at rest, so \(v_{0} = 0.\)

Using the following equation:

\(\Delta x=v_{0}t+\frac{1}{2}at^{2}\\\)

i.e. \(3000 = 0 \times t + \frac{1}{2} \times 3 \times t^{2}\\\)

i.e. \(3000 = \frac{1}{2} \times 3 t^{2}\\\)

i.e. \(t^{2} = 2000\\\)

thus time t= 44.72 sec.

Therefore the required time will be 44.72 sec.

Q.3: A person can travel at a constant velocity of \(11 ms^{-1}\) for 5 minutes time. How far has he travelled?

Solution: At the constant velocity, \(v_{0}= 11 ms^{-1}.\)

Also at time t,\( v =  11 ms^{-1}.\)

The time, t = 5 min = 300 sec

Now we may use the following equation to solve for displacement,

\(\Delta x = (\frac{v+v_{0}}{2})t\\\)

\(\Delta x = (\frac{11 + 11}{2}) \times 300\\\)

\(= 11 \times 300\\\)

= 3300 m.

Therefore, the total displacement will be 3300 m.

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