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The kinematic equations can be referred to those equations that help in understanding the motion of an object with uniform acceleration. They require good knowledge of derivatives, rate of change, and integrals.
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Kinematics Equations
In kinematic equations, we use specific notation to denote initial and final measurements. For example we write \(v\)0 for initial velocity, ? for final velocity, and acceleration (change in velocity) of the object is denoted by \(\Delta \)\(v\) = \(v\) -\(v\)0. This kind of notation also applies to displacement and time. If t0 is initial time, t is final time, then displacement is \(\Delta \)t=t (t0=0 for kinematic equations)
Kinetic equations link five types of variables.
- Displacement (\(\Delta \)x)
- Initial velocity (\(v\)0)
- Final velocity ( \(v\) )
- Time interval (t)
- Constant acceleration (a)
In fact, kinematics equations can define motion at either constant acceleration or constant velocity. We cannot use them if either of the two is changing, as kinematics equations are applicable only at a constant acceleration or a constant speed.
Inverse Kinematics
Inverse kinematics works in the opposite manner in comparison to kinematics. In case there is an endpoint of a particular formation, some of the angle values would be required by the joints in order to achieve that endpoint. It is a little hard and often has infinite or more than one solution. The 4 kinematic equations describing an object's motion are:
- \(v\)= \(v\)0 + at
- \(\Delta \)\(x\) = (\(v\)+\(v\)0/2)t
- \(\Delta \)\(x\) = \(v\)0t + ½ at²
- \(v\)² = \(v\)0² + 2a\(\Delta \)\(x \)
Where d stands for displacement, t stands for the time for which the object moved, a stands for the acceleration of the object, vi stands for the initial velocity of the object, vf stands for the final velocity of the object.
We can notice that if any 4 of the variables are given, we can easily calculate the 5th variable using kinematic equations.
Rotational Kinematic Equations
The above are translational or linear kinematic equations dealing with the motion of a linearly moving body. There is another branch of kinematics equations, known as Rotational Kinematics Equations which deals with the rotational motion of any body. These are however an extension of the above equations with just the variables changed. The differences are given below:
- Instead of displacement, we use change in angle
- Instead of initial and final velocities, we use initial and final angular velocities
- Instead of acceleration, we use angular
- Time is the only constant
Below is the table showing the differences in equations among both rotational motion and linear motion
| Rotational Motion (αα = constant) | Linear Motion (a = constant) |
| ω=ω0+αt | v=v0+at |
| Θ=½(ω+ω0)t | x=½ (v0+v)t |
| Θ=ω0t+½αt2 | x=v0t+½at2 |
| ω2=ω20+2αΘ | v2=v20+2ax |
Derivation of Kinematic Equations
- \(v\) = \(v\)0 + at
Acceleration (a) is defined as the change in velocity (\(\Delta \)v) of an object over the change in time (\(\Delta \)t). It is represented by the equation a = \(\Delta \)v/\(\Delta \)t.
We know that \(\Delta \)v= v – v0
Put this\(\Delta \)v= v – v0 in the acceleration formula
A= v – v0/ \(\Delta \)t.
Simplifying this, we get v= v0+at
- \(\Delta \)\(x \) = [(\(v\)+\(v\)0)/2]t
Consider velocity – time graph with constant acceleration. The slope of velocity touching the y axis is represented as acceleration and the area below the slope is displacement of the object \(\Delta \)x.
Notice here that the height of the blue colored rectangle v0, and the width is t. So the area is equal to v0t (area of rectangle = length x breadth)
The base of triangle is t, the height of the triangle is v-v0.
So the area of the triangle will be ½ t (v-v0)
When we add the areas of rectangle and triangle, we get
\(\Delta \)x = v0t + ½ t (v-v0)
Simplifying we get
\(\Delta \)x = v0t + ½ vt - ½ v0t
Combining initial value terms, we get \(\Delta \)x = v0t + ½ at2
On further simplification, we get\(\Delta \)x = [(v+v0)/2] t
- Third Equation:
By substituting the first kinematics equation into second one, and simplifying we get
The third kinematics equation
\(\Delta \)\(x \) = \(v\)0t + ½ at²
- Fourth Equation:
This equation can be derived using the first and second equations
Solving the first kinematics formula we get t = ((\(v\) = \(v\)0) /a
Substitute this formula in the second equation, we get \(\Delta \)x = [(v+v0)/2] [(v – v0)/a]
Simplifying we get \(\Delta \)x = (\(v\)² - \(v\)0²)/2a
Solving for v2, we get , \(v\)² = \(v\)0² + 2a\(\Delta \)\(x \)
Choosing the Right Equation
The kinematic equations can be applied to a variety of problems related to the motion of an object with constant acceleration. When we solve the problems, the formula we choose should include an unknown variable, as well as 3 known (given) variables. Know that each of the kinematics equations is missing 1 variable. This keen observation helps you to identify what variable is not given or asked for the problem, before selecting an equation that is also missing that variable.
Points to Remember
- Kinematic equation is a part of CBSE class 11 Physics syllabus.
- It comes under Unit 2 Kinematics and carries a total of 10 to 12 periods and 4 to 6 marks.
- Average velocity of an object can be determined by kinematic equations.
- Kinematic equations are used when the acceleration of a body is constant.
- Kinematic equations are a group of four equations which helps in obtaining the unknown information about the motion of an object if other informations are not available.
Sample Questions
Ques 1. How fast will an object moving along the x-axis at t=10 seconds, speed s = 2 m/s, and a constant acceleration of 2 m/s2? (1 mark)
Ans: We know that v= v0+at
Given v0= 2, a= 2, t = 10
So v= 2+2(10) = 22 m/s
Ques 2. A car is moving towards a cliff with an initial speed of 15 m/s. The maximum negative acceleration that the brakes car provides is -0.3 m/s2. If the cliff is 350 meters from the initial position of the car, can the car go over the cliff? (2 marks)
Ans: For the car to stop on the cliff, the final speed must be zero.
V2= v02+2a (x-x0)
0=152+2 (-0.3) (x-0)
= 225+2(-0.3)(x)
=225-0.6 x
=375 m
The final speed is not zero, and so the car cannot stop at the cliff, but goes over the cliff.
Ques 3. Car A moves at a uniform speed past point 1 on a straight track at 0.3 m/s. Cart B moves past point 1 at 0.1 m/s, but is uniformly accelerating at 0.1 m/s2. Point 2 is 1.0 m past point 1. Which cart reaches to point 2 first? Find the time for each cart to reach point 2? (2 marks)
Ans: For the Cart A:
x-x0= v0t+½at2
1.0-0 = 0.3 t + 0
t= 3.3 seconds
For the Cart B:
x-x0= v0t+½at2
1.0-0=0.1t+½ (0.1)t2
Using system of quadratic equation, we get t=3.6 seconds
Therefore Cart A reaches the point 2 first
Ques 4. An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until it finally lifts off the ground. Determine the distance traveled before takeoff. (1 mark)
Ans. a=+3.2m/s²
t = 32.8s
vi = 0 m/s
d = ?
d = vi
d = vi*t + 0.5*a*t²
d = (0 m/s)*(32.8s) + 0.5*(3.20 m/s²)*(32.8s)²
d = 1720 m
Ques 5. A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car. (1 mark)
Ans. d = 110 m
t = 5.21 s
vi = 0 m/s
a = ?
d = vi*t+0.5*a*t²
110 m = (0 m/s)*(5.21s) + 0.5*(a)*(5.21s)²
110 m = (12.57 s²)*a
a = (110m)/(13.57 s²)
a = 8.10 m/s²
Ques 6. Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall? (2 marks)
Ans. a = -9.8 m
t = 2.6 s
vi = 0 m/s
d = ?, vi = ?
d = vi*t + 0.5*a*t²
d = (0 m/s)*(2.60 s) + 0.5*(-9.8 m/s²)*(2.60 s)²
d = -33.1 m (- indicates direction)
vf = vi +a*t
vf = 0 + (-9.8 m/s²)*(2.60 s)
vf = -25.5 m/s (- indicates direction)
Ques 7. A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled.
Ans. vi = 18.5 m/s
vf = 46.1 m/s
t = 2.47 s
d = ?, a = ?
a = (\(\Delta \) v)/t
a = (46.1 m/s - 18.5 m/s)/(2.47 s)
a = 11.2 m/s²
d = vi*t + 0.5*a*t²
d = (18.5 m/s)*(2.47s) + 0.5*(11.2 m/s²)*(2.47s)²
d = 45.7 m + 34.1 m
d = 79.8 m
Ques 8. A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s². Determine the time for the feather to fall to the surface of the moon. (1 mark)
Ans. vi = 0 m/s
d = -1.40 m
a = -1.67 m/s
t = ?
d = vi*t + 0.5*a*t²
-1.40 m = (0 m/s)*(t)+0.5*(-1.67 m/s²)*(t)²
-1.40 m = 0 + (-0.835 m/s²)*(t)²
(-1.40m)/(-0.835 m/s²) = t²
1.68s² = t²
t = 1.29 s
Ques 9. Rocket-powered sleds are used to test the human response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 m/s in 1.83 seconds, then what is the acceleration and what is the distance that the sled travels? (2 marks)
Ans. vi = 0 m/s
vf = 444 m/s
t = 1.83 s
a = ?, d = ?
a = (Delta v)/t
a = (444 m/s - 0 m/s)/(1.83 s)
a = 243 m/s²
d = vi*t + 0.5*a*t²
d = (0 m/s)*(1.83 s) + 0.5*(243 m/s²)*(1.83 s)²
d = 0 m + 406 m
d = 406 m
Ques 10. A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike. (1 mark)
Ans. vi = 0 m/s
vf = 7.10 m/s
d = 35.4 m
a = ?
vf² = vi² + 2ad
(7.10 m/s)² = (0 m/s)² + 2a(35.4 m)
50.4m²/s² = (0 m/s)² + (70.8 m)a
(50.4m²/s²)/(70.8m) = a
a = 0.712 m/s²
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