Kinematics Formula: Definition, Derivation & Example

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Kinematics is the study of motion of any particular point or object with respect to space and time. It is a branch of classical mechanics. Kinematics deals with the study of motion of any points, objects and groups of objects while ignoring the causes of motion. Kinematics is also referred to as “Geometry of Motion” by many physicists.

Kinetic Energy of an object is the energy that the body possesses due to its motion. It is defined as the work needed to accelerate a body of given mass from rest to a stated velocity.

In order to describe motion, Kinematics focuses on various geometric objects and differential properties including velocity and acceleration. Kinematics is used in astrophysics, mechanical engineering, robotics and biomechanics.

Kinematic equations are a set of equations or formula that derive one of the five kinematic variables namely,

  • Displacement (D),
  • Initial Velocity (u),
  • Final Velocity (v),
  • Time Interval (t),
  • Constant Acceleration (a)

when other variables are provided. These equations define motion either at constant velocity or at constant acceleration. Since such is the case, we cannot use the equations if either of the two is variable or changing.

Read more: Period Motion

KeyTerms: Kinematic, Velocity, Acceleration, Displacement, Time, Mass, Motion.


Kinematic Formulas

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The Kinematic Equations of motion are as follows:

  1. v = u + at
  2. v2 = u2 + 2aD
  3. D = ut + \(\frac{at^2}{2}\)
  4. D = \((\frac{v + u}{2}) t\)

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Kinematic Formula Derivations

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Formula 1: v = u + at

Derivation:

We know that Acceleration of a body is the rate of change of velocity of that body.

Now, let us assume that a body has mass “m” and initial velocity “u”. After some time “t” its velocity changes to “v” which is its final velocity due to uniform acceleration “a”.

We know that,

\(Acceleration = \frac{Final Velocity - Initial Velocity}{Time Taken}\)

Or, a = \(\frac{v - u}{t}\)

Or, at = v - u

Therefore, v = u + at [Derived]

Formula 2: v2 = u2 + 2aD

Derivation:

Let us assume, Displacement be “D”

From the first equation, we know that, v= u + at . Therefore, we can write it as, 

t=(v - u)/ a

Also, we know that, Displacement= Average Velocity Time

Hence, for constant acceleration we can write,

Average Velocity = (Final Velocity + Initial Velocity)/2

= (v + u)/2

Therefore, Displacement D= \([\frac{v + u}{2}] \times [\frac{v-u}{a}]\)

Or, D = (v2 – u2)/2a

Or, 2aD = v2 – u2

Or, v2 = u2 + 2aD [Derived]

Formula 3: D = ut + \(\frac{at^2}{2}\)

Derivation:

We know that,

Displacement = Average Velocity x Time

Also,

Average Velocity = (u+v) / 2

Displacement D = {\({\frac{u+v}{2}}\)} x t

From the first equation = u + at ,

We get,

D = \(\frac{u + u + at}{2} \times t\)

Or, D = \(\frac{2u + at}{2} \times t\)

Or, D = (2ut + at2) / 2

Or, D = \(\frac{2ut}{2} + at^2/2\)

Therefore, D = ut + \(\frac{at^2}{2}\) [Derived]

Formula 4: D = \((\frac{v + u}{2})t\)

Derivation:

Derivation

Consider the above 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 “D”.

The base of triangle is “t” the height of the triangle is “v – u" 

So, the area of the triangle will be = ½ t (v – u)

When we add the areas of rectangle and triangle, we get,

 D = ut + ½ t (v – u)

Simplifying this we get,

 D= ut + ½ vt – ½ ut

Combining initial value terms, we get,

 D = ut + ½ at2

On further simplification, we get,

D= \((\frac{v+u}{2})t\) [Derived]


Rotational Kinematics

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There is another branch of kinematics equations which deals with the rotational motion of a point or an object besides the Transitional or Linear Kinematic equations as discussed earlier that involves kinematics of bodies in linear motion. Displacement is replaced by a change in angle.

  • Initial and final velocities are replaced by initial and final angular velocity.
  • Acceleration is replaced by angular acceleration
  • Time is the only constant

Rotation Kinematics 

Rotation Kinematics 

The equations of Rotational Kinematics are:

ω = ω0 + αt

θ = ω + \(\frac{(\omega + \omega _0t)}{2}\)

θ = ω0 t + \(\frac{(at^2)}{2}\)

ω2 = ω02 + 2α θ


Things to remember

  • The point or object always moves in a straight line and in the same direction.
  • The magnitude of velocity is equal to speed.
  • The average velocity is equal to instantaneous velocity.
  • Since velocity is constant, acceleration is 0 and hence net force or resultant force is also 0.
  • Acceleration is a vector quantity.
  • If the value of acceleration is negative, it is called Retardation.
  • Velocity is a vector quantity and speed is a scalar quantity.
  • Units of speed and velocity are “meter per second” or “ms-1”.

Sample Question

Ques. What is Kinematics? (2 marks)

Ans. The branch of physics that defines motion with respect to space and time, ignoring the cause of that motion, is known as kinematics. Kinematics equations are a set of equations that can derive an unknown aspect of a body’s motion if the other aspects are provided.

Ques. What is Kinetic Energy? (1 mark)

Ans. The kinetic energy is the measure of the work an object can do by virtue of its motion.

Ques. 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? (2 marks)

Ans. Known v= u+at

Given u= 2, a= 2, t = 10

So v= 2+2(10) = 22 m/s

Ques. 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 0.

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. 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? (5 marks)

Ans. For the Cart A: 

x – x0 = v0t + ½ at2 

1.0 – 0 = 0.3 t + 0 

t= 3.3 seconds

For the Cart B:

x – x0 = v0t + ½ at2 

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. 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. (5 marks)

Ans. a = + 3.2 m/s2

t = 32.8s

vi = 0 m/s

d = ?

d = vi

d = vit + 0.5*a*t2

d = (0 m/s)*(32.8s) + 0.5*(3.20 m/s²)*(32.8s)2

d = 1720 m

Ques. 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. (4 marks)

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. 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? (5 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. 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. (5 marks)

Ans. vi = 18.5 m/s 

vf = 46.1 m/s

t = 2.47 s

d = ?, a = ?

a = (Δv)/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. 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. (5 marks)

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. 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, what is the acceleration and the distance that the sled travels? (5 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. 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. (4 marks)

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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