Kinematics Equations
The kinematics equations in physics conveniently calculate the distance, velocity, and acceleration of 1D or 2D motions. It is essential to know them and how to use them for the first exam of the first course of physics at the college. See the examples below for reference.
\begin{array}{{>{\displaystyle}l}}
v=v0+at\\
v^{2} -v_{0}^{2} =2a( x-x_{0})\\
x-x_{0} =v_{0} t+\frac{1}{2} at^{2}\\
x-x_{0} =\frac{1}{2}( v_{0} +v) t\\
x-x_{0} =vt-\frac{1}{2} at^{2}
\end{array}
\begin{array}{{>{\displaystyle}l}}
x_{0} =initial\ position\ ( m)\\
x=final\ position\ ( m)\\
v_{0} =initial\ velocity\ ( m/s)\\
v=final\ velocity\ ( m/s)\\
a=acceleration\ ( constant) \ \left( m/s^{2}\right)\\
t=time\ ( duration) \ ( sec)
\end{array}
The relationship between position, velocity, and acceleration is as follows. The kinematics equations are derived from this concept.
f(x) = position
f'(x) = velocity (first derivative of position)
f”(x) = acceleration (second derivative of position)
Kinematics Equations Example 1
You are driving a car at 34 miles/h and see the light is being changed to yellow. The maximum deceleration of your car is 3.2 m/s². Should you brake your car to attempt to stop without entering the intersection or maintain the current speed and drive through the intersection without getting a red light violation? The duration of the yellow light is 2.5 seconds, and the distance between your car’s position when the light is changed to yellow and the white stop line at the intersection is 40 m.
Solution
Let’s convert the car’s velocity to the SI unit.
\begin{equation*}
\left(\frac{34\ miles}{h}\right)\left(\frac{1609.34\ m}{1\ mile}\right)\left(\frac{1\ hr}{3600\ s}\right) =15.2\ m/s
\end{equation*}
If you don’t step on the brake
\begin{equation*}
\left(\frac{15.20\ m}{s}\right)( 2.5\ s) =38.0\ m\
\end{equation*}
You can only drive 38.0 m during the 2.5-second yellow light duration. This is short of 40 m. Therefore, you will get a red light violation.
If you try to stop your car,
The deceleration is a negative acceleration, and the distance between the initial point and the final point is 40 m.
a = -3.2 m/s²
dx = (x-x0) = (40 m – 0 m)
v0 = 34 miles/h
v = 0 m/s (final velocity)
and use one of the above equations,
\begin{gather*}
v^{2} -v_{0}^{2} =2a( x-x_{0})\\
\\
\frac{v^{2} -v_{0}^{2}}{2a} =x-x_{0}\\
\\
x-x_{0} =\frac{v^{2} -v_{0}^{2}}{2a}\\
\\
x-x_{0} =\frac{\left(\frac{0\ m}{s}\right)^{2} -\left(\frac{15.20\ m}{s}\right)^{2}}{2\left( -\frac{3.2\ m}{s^{2}}\right)} =36.1\ m
\end{gather*}
It takes 36.1 m to decelerate and stop. You will be able to stop without entering the intersection.
Kinematics Equations Example 2
The basketball is thrown vertically above the ground. The initial velocity is 35 m/s straight upward. (a) How long does it take to reach half the velocity of the initial velocity? (b) How far does it travel from the ground?
Solution
We need to define which direction is positive. Let’s say upward is positive. The gravitational acceleration is negative, thus -9.8 m/s.
Part (a)
\begin{gather*}
( a) \ Use\ one\ of\ equations\ above\\
v=v_{0} +at\\
\\
v-v_{0} =at\\
\\
\frac{v-v_{0}}{a} =t\\
\\
t=\frac{v-v_{0}}{a}\\
\\
v\ =\ half\ the\ velocity\ of\ initial\ velocity\\
v=\frac{1}{2} v_{0}\\
\\
plug\ this\ one\ into\ above\ equation\\
t=\frac{\frac{1}{2} v_{0} -v_{0}}{a}\\
\\
t=\frac{-\frac{1}{2} v_{0}}{a}\\
\\
a=-9.8\ m/s\ gravitational\ acceleration.\\
it’s\ negative\ because\ it’s\ downward\\
\\
t=\frac{\left( -\frac{1}{2}\right)\left(\frac{35\ m}{s}\right)}{\left(\frac{-9.8\ m}{s^{2}}\right)} =1.79\ s\\
\\
\\
\\
part\ ( b)\\
\\
x_{0} \ =\ 0\ m\\
dx=x-x_{0} =x-0\ m=x\\
\\
use\ one\ of\ above\ equations\\
\\
x-x_{0} =\frac{1}{2}( v_{0} +v) t\\
\\
x-x_{0} =\frac{1}{2}\left(\frac{35\ m}{s} +\left(\frac{1}{2}\right)\left(\frac{35\ m}{s}\right)\right)( 1.79\ s) =46.99\ m\\
\\
Alternatively,\\
You\ can\ use\ another\ equation\\
\\
x-x_{0} =v_{0} t+\frac{1}{2} at^{2}\\
\\
x-x_{0} =\left(\frac{35\ m}{s}\right)( 1.79\ s) +\frac{1}{2}\left(\frac{-9.8\ m}{s^{2}}\right)( 1.79\ s)^{2} =46.99\ m\\
\\
Also,\ you\ can\ this\ equation\ too\\
x-x_{0} =vt-\frac{1}{2} at^{2}\\
\\
x-x_{0} =\left(\frac{1}{2}\right)\left(\frac{35\ m}{s}\right)( 1.79\ s) -\frac{1}{2}\left(\frac{-9.8\ m}{s^{2}}\right)( 1.79\ s)^{2} =46.99\ m
\end{gather*}
Conclusion. It takes 1.79 seconds to reach half the initial velocity at 46.99 m high.
The maximum height and the time it takes to reach the maximum height can be calculated by setting v=0 m/s at the maximum height, which is x.
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