In Physics, Motion is the state of body in which it changes its position with time. Motion is fundamentally described by physical quantities such distance, displacement, speed, velocity, acceleration, and time. These physical quantities can be expressed in the form of a mathematical equation to express motion. These equations are called Equations of Motion. These equations can be derived via various methods such as Algebraic Method, Graphical Method and Calculus Method.
This article deals with the equation of motion and its derivation using the calculus method. This derivation is useful for class 11 students.
Fundamentals of Motion
The fundamentals of motion in physics are described by the concepts: of distance, displacement, speed, velocity, acceleration, and time. Here’s a brief overview:
Distance and Displacement
- Distance is the total length of the path travelled by an object, irrespective of direction. It is a scalar quantity.
- Displacement(s) is the change in position of an object, measured in a straight line from the starting point to the ending point. It is a vector quantity.
Speed and Velocity
- Speed is the rate at which an object covers distance. It is a scalar quantity and is calculated as the distance travelled divided by the time taken.
- Velocity(v) is the rate at which an object changes its position in a particular direction. It is a vector quantity and is calculated as the displacement divided by the time taken.
Acceleration
- Acceleration(a) is the rate at which velocity changes over time. Acceleration is also a vector quantity and is calculated as the change in velocity divided by the time taken.
Time
- Time(t) is a fundamental parameter used to measure the duration of an event or the interval between two events.
Equations of Motion by Calculus Method
There are three equations of motions also called Newton’s Equation of Motion. The expression of the three equations of motions are:
- First Equation of Motion: v = u + at
- Second Equation of Motion: s = ut + 1/2at2
- Third Equation of Motion: v2 – u2 = 2as
In the above three equations, v is final velocity, u is initial velocity, a is acceleration, s is distance and t is time.
Now, let’s derive these equation by calculus method:
First Equation of Motion by Calculus
Newton’s first equation of motion states that final velocity is equal to sum of initial velocity and product of acceleration and time. Mathematically, first equation is expressed as
v = u + at
Mathematically, acceleration is defined as
a = dv / dt
Integrating both sides of the equation with respect to time gives us:
v∫u dv= t∫0 a dt
Where v is the final velocity, u is the initial velocity, a is the constant acceleration, and t is the time.
⇒ v∫u dv= at∫0 dt
⇒ v – u = at
Adding u on both sides, we get the final form
v = u + at
Second Equation of Motion by Calculus
Newton’s second equation of motion states that total distance covered is equal to sum of product of initial velocity and time and half of product of acceleration and square of time. Second equation of motion is mathematically expressed as s = ut + 1/2at2
Now, velocity is rate of change of distance. Hence, in terms of differentiation it can be expressed as
v = ds / dt
Integrating both sides of the equation with respect to time gives us:
s∫0 ds= t∫0 v dt
Now using the first equation of motion
s∫0 ds= t∫0 (u+ at) dt
Where v is the final velocity, u is the initial velocity, s is the displacement, a is the constant acceleration, and t is the time.
s∫0 ds = t∫0 u dt + t∫0 at dt
Since, u and a are constant
⇒ s∫0 ds = ut∫0 dt + at∫0 t dt
⇒ s = ut + at2/2
Hence, we derive the second equation of motion
s = ut+1/2 at2
Third Equation of Motion by Calculus
Third equation of motion states that difference between square of final velocity and initial velocity is equal to the twice the product of acceleration and distance. Mathematically, third equation of motion is given as: v2 – u2 = 2as
We start with the mathematical definition of acceleration
a = dv/dt
Multiply with v on both sides
av = v dv/dt
Substituting the definition of v = ds/dt
a ds/dt = v dv/dt
The time dependence can be removed here and this becomes
a ds = v dv
Integrating both sides
s∫0 a ds= v∫u v dv
Where v is the final velocity, u is the initial velocity, s is the displacement, and a is the constant acceleration.
as = [v2/2]uv
(v2 – u2)/2 = as
⇒ v2 – u2 = 2as
Rearranging, we get the Third Equation of Motion
v2 = u2+2as
Applications of Calculus in Motion
Here are some applications of calculus in motion:
Analysing Uniformly Accelerated Motion
Calculus is used to derive the equations of motion for uniformly accelerated motion. It also enables us to determine instantaneous velocity and acceleration at any given point in time during the motion. By taking derivatives of displacement with respect to time, we obtain velocity functions. Similarly, taking derivatives of velocity with respect to time gives acceleration functions. Calculus allows us to understand how velocity and acceleration change over time. It helps in determining maximum velocity, maximum displacement, and time taken to reach specific positions.
Calculating Displacement and Velocity in Variable Accelerations
In cases of variable acceleration, calculus plays a crucial role in calculating displacement and velocity accurately. The velocity of the object at any time t is the integral of its acceleration function from the initial time t0 to t:
v(t) = t∫t0 a(t)dt
The displacement of the object from the initial position at time t is the integral of its velocity function from the initial time t0 to t:
s(t) = t∫t0 v(t)dt
Next step is to evaluate the integrals, and apply initial conditions to determine constants of integration. This would give the final form required.
Examples on Equation of Motion by Calculus Method
Example 1: A body moving along the x-axis has its position given by x(t)=t3-2t+1. Find its velocity and acceleration as a function of time.
Solution:
The velocity can be obtained by differentiating the position w.r.t. time.
v(t) = dx/dt = 3t2-2
The acceleration can be obtained by differentiating the velocity w.r.t. time.
a(t) = dv/dt = 6t
Example 2: For the above problem find the velocity and acceleration at t = 3s.
Solution:
The velocity :
v(t) = 3t2– 2 = 27 – 2 = 25 m/s
The acceleration can be calculated as:
a(t) = 6t = 18 m/s2
Example 3: A body moving along the x-axis has its velocity given by v(t) = t3-2t+1. Find its displacement between t = 0 to 5s.
Solution:
The displacement is given by integrating the velocity
S= 5∫0 v(t) dt= 5∫0 (t3-2t+1) dt= [t4/4 – t2 + t]05 = [(625/4)- 25 + 5]= 136.25 m
Example 4: A body has acceleration varying as a(t) = 6-t. Find its displacement in t = 0 → 3.5s
Solution:
The displacement is given by integrating the velocity and the velocity is obtained by integrating the acceleration
v(t)= t∫0 a(t) dt= t ∫0 (6-t) dt= 6t-t2/2
S= 3.5∫0 v(t) dt= 3.5∫0 (6t-t2/2)dt = [6t2-t3/6]03.5 = 6 x (49/4) – (343/48)= 66.35 m
Therefore, the total displacement is 66.35 m.
Practice Problems on Equation of Motion by Calculus Method
Q1. A car moves along a straight road with its position given by s(t)=2t2-5t+3, where s is in metres and t is in seconds. Determine the car’s velocity and acceleration at time t =3s.
Q2. The height of a ball thrown upwards is given by h(t)=3.5t-4.9t2, find its velocity and acceleration as a function of time. Also, find the maximum height reached by the ball.
Q3. A particle is moving along the y-axis, and its velocity is given by v(t)= 2t2-3t+5 m/s. Find its displacement between t =1 second and t =4 seconds.
Q4. An object is dropped from a height of 100 metres. Its velocity function as it falls is given by v(t)=9.8t, where t is the time in seconds and v is in metres per second. Determine the height of the object after the first 5 seconds of its fall.
Q5. A car accelerates with an acceleration function a(t)=2t2−3 m/s². Calculate the displacement of the car from t=0 to t=6 seconds.
Conclusion: Equation of Motion by Calculus Method
Calculus plays a vital role in motion analysis by enabling the derivation of Newton’s equations of motion. Additionally, it provides the means to address scenarios of variable accelerations or variable velocity. Calculus allows Newton’s equations to be used for non-linear motion or variable acceleration.
FAQs on Equation of Motion by Calculus Method
How many Equations of Motion are there?
There are three equations of motion:
- First Equation of Motion: v = u + at
- Second Equation of Motion: s = ut + 1/2at2
- Third Equation of Motion: v2 – u2 = 2as
What are other ways to derive equation of motion?
The other methods to derive equation of motion includes graphical method and algebraic method
Why is Calculus Important in the Study of Motion?
Calculus is important in motion analysis as it leads to the derivation of Newton’s equations of motion. Further, it allows us to solve cases involving variable accelerations or time-dependent changes in velocity.
Can Newton’s Equations Be Used for Non-Linear Motion?
Yes, Newton’s equations can be used for non-linear motion or variable acceleration. But that involves calculus and applying integration in equations of non-uniform acceleration.
What are the limitations of classical equations of motion?
These equations of motion do not take into account the theory of relativity, and can be inaccurate for speeds comparable to the speed of light. At very small scales, these become invalid and quantum mechanics comes into play.
How to express velocity in differential form?
Velocity is rate of change of displacement. In differential form it is expressed as v = ds/dt
How to express acceleration in differential form?
Acceleration is rate of change of velocity. In differential form it is expressed as a = dv/dt
Share your thoughts in the comments
Please Login to comment...